Session 09-02: Tasks

Differential Calculus - Exam Review

Differential Calculus - Exam Review

Problem 1: Limit Evaluation (x)

Evaluate the following limits:

1. \(\lim_{x \to 0} \frac{e^{5x}-1}{x}\)

2. \(\lim_{n \to \infty} \left(1+\frac{3}{n}\right)^n\)

3. \(\lim_{x \to \infty} \frac{4x^2-x+3}{2x^2+5}\)

4. \(\lim_{x \to 2} \frac{x^2-4}{x-2}\)

CautionSolution

a) We use the standard limit \(\lim_{x \to 0} \frac{e^{kx}-1}{x} = k\).

Here \(k = 5\), so:

\[\lim_{x \to 0} \frac{e^{5x}-1}{x} = 5\]

Alternatively, applying L’Hôpital’s rule since \(\frac{0}{0}\):

\[\lim_{x \to 0} \frac{5e^{5x}}{1} = 5 \cdot e^0 = 5\]

b) We use the standard limit \(\lim_{n \to \infty} \left(1+\frac{k}{n}\right)^n = e^k\).

Here \(k = 3\), so:

\[\lim_{n \to \infty} \left(1+\frac{3}{n}\right)^n = e^3\]

c) Divide numerator and denominator by \(x^2\) (the highest power):

\[\lim_{x \to \infty} \frac{4x^2-x+3}{2x^2+5} = \lim_{x \to \infty} \frac{4 - \frac{1}{x} + \frac{3}{x^2}}{2 + \frac{5}{x^2}} = \frac{4-0+0}{2+0} = 2\]

d) Factor the numerator:

\[\lim_{x \to 2} \frac{x^2-4}{x-2} = \lim_{x \to 2} \frac{(x-2)(x+2)}{x-2} = \lim_{x \to 2} (x+2) = 4\]

Problem 2: Differentiation Drill (x)

Find \(f'(x)\) for each function:

1. \(f(x) = 3x^4 - 2x^3 + 7x - 1\)

2. \(f(x) = e^{3x+1}\)

3. \(f(x) = \ln(2x+5)\)

4. \(f(x) = \frac{x^2+1}{x}\)

5. \(f(x) = (3x-1)^5\)

6. \(f(x) = x \cdot e^x\)

CautionSolution

a) Apply the power rule term by term:

\[f'(x) = 12x^3 - 6x^2 + 7\]

b) Apply the chain rule with the inner function \(u = 3x+1\), so \(u' = 3\):

\[f'(x) = e^{3x+1} \cdot 3 = 3e^{3x+1}\]

c) Apply the chain rule with the inner function \(u = 2x+5\), so \(u' = 2\):

\[f'(x) = \frac{1}{2x+5} \cdot 2 = \frac{2}{2x+5}\]

d) Rewrite first: \(f(x) = \frac{x^2+1}{x} = x + \frac{1}{x} = x + x^{-1}\)

\[f'(x) = 1 - x^{-2} = 1 - \frac{1}{x^2}\]

Alternatively, using the quotient rule:

\[f'(x) = \frac{2x \cdot x - (x^2+1) \cdot 1}{x^2} = \frac{2x^2 - x^2 - 1}{x^2} = \frac{x^2 - 1}{x^2} = 1 - \frac{1}{x^2}\]

e) Apply the chain rule with the inner function \(u = 3x-1\), so \(u' = 3\):

\[f'(x) = 5(3x-1)^4 \cdot 3 = 15(3x-1)^4\]

f) Apply the product rule with \(u = x\) and \(v = e^x\):

\[f'(x) = 1 \cdot e^x + x \cdot e^x = e^x(1+x)\]

Problem 3: Basic Critical Points (x)

Find all critical points and classify them using the second derivative test:

1. \(f(x) = x^3 - 12x + 5\)

2. \(f(x) = -2x^2 + 8x - 3\)

CautionSolution

a) \(f(x) = x^3 - 12x + 5\)

Step 1: Find \(f'(x)\):

\[f'(x) = 3x^2 - 12\]

Step 2: Set \(f'(x) = 0\):

\[3x^2 - 12 = 0 \implies x^2 = 4 \implies x = \pm 2\]

Step 3: Find \(f''(x)\):

\[f''(x) = 6x\]

Step 4: Apply the second derivative test:

• At \(x = -2\): \(f''(-2) = -12 < 0\) \(\Rightarrow\) local maximum at \((-2, f(-2)) = (-2, 21)\)
  • \(f(-2) = (-2)^3 - 12(-2) + 5 = -8 + 24 + 5 = 21\)
• At \(x = 2\): \(f''(2) = 12 > 0\) \(\Rightarrow\) local minimum at \((2, f(2)) = (2, -11)\)
  • \(f(2) = 8 - 24 + 5 = -11\)

b) \(f(x) = -2x^2 + 8x - 3\)

Step 1: Find \(f'(x)\):

\[f'(x) = -4x + 8\]

Step 2: Set \(f'(x) = 0\):

\[-4x + 8 = 0 \implies x = 2\]

Step 3: Find \(f''(x)\):

\[f''(x) = -4\]

Step 4: Apply the second derivative test:

• At \(x = 2\): \(f''(2) = -4 < 0\) \(\Rightarrow\) local maximum at \((2, f(2)) = (2, 5)\)
  • \(f(2) = -2(4) + 8(2) - 3 = -8 + 16 - 3 = 5\)

Since this is a downward-opening parabola, this local maximum is also the global maximum.

Problem 4: Critical Point Classification (xx)

For \(f(x) = x^4 - 8x^2 + 16\):

1. Find all critical points.

2. Use the second derivative test to classify them. Handle the case where \(f''(x_0) = 0\) carefully.

3. Find the global extrema on the interval \([-3, 3]\).

4. Determine the intervals of increase and decrease.

CautionSolution

a) Find \(f'(x)\):

\[f'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x-2)(x+2)\]

Setting \(f'(x) = 0\): \(x = 0, \; x = 2, \; x = -2\)

b) Find \(f''(x)\):

\[f''(x) = 12x^2 - 16\]

Apply the second derivative test:

• At \(x = -2\): \(f''(-2) = 12(4) - 16 = 32 > 0\) \(\Rightarrow\) local minimum
  • \(f(-2) = 16 - 32 + 16 = 0\)
• At \(x = 0\): \(f''(0) = -16 < 0\) \(\Rightarrow\) local maximum
  • \(f(0) = 0 - 0 + 16 = 16\)
• At \(x = 2\): \(f''(2) = 12(4) - 16 = 32 > 0\) \(\Rightarrow\) local minimum
  • \(f(2) = 16 - 32 + 16 = 0\)

Note: In this problem the second derivative test works for all critical points. If \(f''(x_0) = 0\) had occurred, we would need to examine the sign change of \(f'(x)\) around \(x_0\) directly.

c) Global extrema on \([-3, 3]\):

Evaluate \(f\) at the critical points and endpoints:

• \(f(-3) = 81 - 72 + 16 = 25\)
• \(f(-2) = 0\)
• \(f(0) = 16\)
• \(f(2) = 0\)
• \(f(3) = 81 - 72 + 16 = 25\)

Global maximum: \(f(-3) = f(3) = 25\) (at the endpoints)

Global minimum: \(f(-2) = f(2) = 0\) (at the local minima)

d) Sign analysis of \(f'(x) = 4x(x-2)(x+2)\):

Interval	\(4x\)	\(x-2\)	\(x+2\)	\(f'(x)\)	Behavior
\((-\infty, -2)\)	\(-\)	\(-\)	\(-\)	\(-\)	decreasing
\((-2, 0)\)	\(-\)	\(-\)	\(+\)	\(+\)	increasing
\((0, 2)\)	\(+\)	\(-\)	\(+\)	\(-\)	decreasing
\((2, \infty)\)	\(+\)	\(+\)	\(+\)	\(+\)	increasing

Problem 5: Quadratic Optimization Word Problem (xx)

A rectangular garden is to be fenced on three sides (the fourth side is along a wall). The total available fencing is 120 meters.

1. Express the area as a function of one variable.

2. Find the dimensions that maximize the area.

3. What is the maximum area?

4. Verify using the second derivative test.

CautionSolution

a) Let \(x\) be the length of the side parallel to the wall and \(y\) be the length of each perpendicular side. The fencing constraint is:

\[x + 2y = 120 \implies x = 120 - 2y\]

The area is:

\[A(y) = x \cdot y = (120 - 2y) \cdot y = 120y - 2y^2\]

with the constraint \(0 < y < 60\) (so that both \(x > 0\) and \(y > 0\)).

b) Find the critical point:

\[A'(y) = 120 - 4y = 0 \implies y = 30\]

Then \(x = 120 - 2(30) = 60\).

The optimal dimensions are \(x = 60\) m (parallel to wall) and \(y = 30\) m (perpendicular to wall).

c) The maximum area is:

\[A(30) = 120(30) - 2(30)^2 = 3600 - 1800 = 1800 \text{ m}^2\]

d) Verification with the second derivative test:

\[A''(y) = -4 < 0 \text{ for all } y\]

Since \(A''(30) = -4 < 0\), the critical point \(y = 30\) is indeed a maximum.

Interpretation: To maximize the garden area with 120 m of fencing along three sides, the garden should be 60 m wide (along the wall) and 30 m deep, yielding a maximum area of 1800 m².

Problem 6: Polynomial Curve Sketch (xx)

For \(f(x) = x^3 - 3x^2 - 9x + 27\):

1. Find the zeros (factor the polynomial).

2. Find the critical points and classify them.

3. Find the inflection points.

4. Determine the end behavior.

5. Sketch the complete graph.

CautionSolution

a) Try \(x = 3\): \(f(3) = 27 - 27 - 27 + 27 = 0\). So \((x-3)\) is a factor.

Perform polynomial division:

\[f(x) = (x-3)(x^2 - 9) = (x-3)(x-3)(x+3) = (x-3)^2(x+3)\]

Zeros: \(x = 3\) (double zero) and \(x = -3\) (simple zero).

b) Find \(f'(x)\):

\[f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1)\]

Critical points: \(x = 3\) and \(x = -1\).

Find \(f''(x)\):

\[f''(x) = 6x - 6\]

Classification:

• At \(x = -1\): \(f''(-1) = -6 - 6 = -12 < 0\) \(\Rightarrow\) local maximum
  • \(f(-1) = -1 - 3 + 9 + 27 = 32\)
• At \(x = 3\): \(f''(3) = 18 - 6 = 12 > 0\) \(\Rightarrow\) local minimum
  • \(f(3) = 0\) (the double zero)

c) Set \(f''(x) = 0\):

\[6x - 6 = 0 \implies x = 1\]

Check sign change: \(f''(0) = -6 < 0\) and \(f''(2) = 6 > 0\). Sign change confirmed.

Inflection point: \((1, f(1)) = (1, 1 - 3 - 9 + 27) = (1, 16)\)

d) End behavior (leading term \(x^3\)):

• As \(x \to \infty\): \(f(x) \to \infty\)
• As \(x \to -\infty\): \(f(x) \to -\infty\)

e) Key features for the sketch:

• Zeros at \(x = -3\) (crosses x-axis) and \(x = 3\) (touches x-axis, double root)
• Local maximum at \((-1, 32)\)
• Local minimum at \((3, 0)\)
• Inflection point at \((1, 16)\)
• \(y\)-intercept: \(f(0) = 27\)
• Rises to the right, falls to the left

Problem 7: Tangent Line at a Point (xx)

Consider the function \(f(x) = x^2 \cdot e^{-x}\).

1. Find the equation of the tangent line to \(f\) at \(x = 2\).

2. Find the equation of the normal line at the same point.

3. At which point(s) on the curve is the tangent line horizontal?

CautionSolution

a) First, find \(f(2)\):

\[f(2) = 4 \cdot e^{-2} = \frac{4}{e^2}\]

Next, find \(f'(x)\) using the product rule:

\[f'(x) = 2x \cdot e^{-x} + x^2 \cdot (-e^{-x}) = e^{-x}(2x - x^2) = x(2-x)e^{-x}\]

The slope at \(x = 2\):

\[f'(2) = 2(2-2)e^{-2} = 0\]

The tangent line at \(x = 2\) is horizontal:

\[y = \frac{4}{e^2} \approx 0.541\]

b) Since the tangent line is horizontal (slope \(= 0\)), the normal line is vertical:

\[x = 2\]

c) The tangent line is horizontal where \(f'(x) = 0\):

\[x(2-x)e^{-x} = 0\]

Since \(e^{-x} \neq 0\) for all \(x\), we need \(x(2-x) = 0\), giving \(x = 0\) or \(x = 2\).

The points with horizontal tangent lines are:

• \((0, f(0)) = (0, 0)\)
• \((2, f(2)) = \left(2, \frac{4}{e^2}\right) \approx (2, 0.541)\)

Problem 8: Rational Curve Sketch with Asymptotes (xxx)

For \(f(x) = \frac{x^2 - 2x - 3}{x - 2}\):

1. Determine the domain.

2. Find the vertical asymptote(s).

3. Perform polynomial long division to find the oblique asymptote.

4. Find all zeros and the \(y\)-intercept.

5. Compute \(f'(x)\) and find the critical points.

6. Compute \(f''(x)\) and find the inflection points.

7. Sketch the complete graph.

CautionSolution

a) The denominator cannot be zero: \(x - 2 \neq 0 \implies x \neq 2\).

\[D_f = \mathbb{R} \setminus \{2\}\]

b) Vertical asymptote at \(x = 2\).

Check: \(f(2)\) is undefined, and the numerator at \(x = 2\): \(4 - 4 - 3 = -3 \neq 0\). So this is indeed a vertical asymptote (not a removable discontinuity).

Behavior near \(x = 2\):

• As \(x \to 2^+\): numerator \(\to -3\), denominator \(\to 0^+\), so \(f(x) \to -\infty\)
• As \(x \to 2^-\): numerator \(\to -3\), denominator \(\to 0^-\), so \(f(x) \to +\infty\)

c) Polynomial long division of \((x^2 - 2x - 3) \div (x - 2)\):

\[x^2 - 2x - 3 = (x-2)(x) + (-3)\]

Wait, let’s verify: \((x-2) \cdot x = x^2 - 2x\). Remainder: \((x^2 - 2x - 3) - (x^2 - 2x) = -3\).

\[f(x) = x + \frac{-3}{x-2} = x - \frac{3}{x-2}\]

The oblique asymptote is \(y = x\).

d) Zeros: Set \(x^2 - 2x - 3 = 0\):

\[(x-3)(x+1) = 0 \implies x = 3 \text{ or } x = -1\]

Both are in the domain.

\(y\)-intercept: \(f(0) = \frac{0 - 0 - 3}{0 - 2} = \frac{-3}{-2} = \frac{3}{2}\)

e) Using \(f(x) = x - \frac{3}{x-2}\):

\[f'(x) = 1 + \frac{3}{(x-2)^2}\]

Set \(f'(x) = 0\):

\[1 + \frac{3}{(x-2)^2} = 0 \implies \frac{3}{(x-2)^2} = -1\]

This has no solution since \((x-2)^2 > 0\), so \(\frac{3}{(x-2)^2} > 0\) always.

Therefore \(f'(x) > 0\) for all \(x \in D_f\), and there are no critical points. The function is strictly increasing on each branch of its domain.

f) Find \(f''(x)\):

\[f'(x) = 1 + 3(x-2)^{-2}\]

\[f''(x) = -6(x-2)^{-3} = \frac{-6}{(x-2)^3}\]

Set \(f''(x) = 0\): \(-6 = 0\) has no solution. There are no inflection points.

However, \(f''(x)\) changes sign at \(x = 2\) (the asymptote):

• For \(x < 2\): \((x-2)^3 < 0\), so \(f''(x) = \frac{-6}{< 0} > 0\) \(\Rightarrow\) concave up
• For \(x > 2\): \((x-2)^3 > 0\), so \(f''(x) = \frac{-6}{> 0} < 0\) \(\Rightarrow\) concave down

g) Key features for the sketch:

• Domain: \(\mathbb{R} \setminus \{2\}\)
• Vertical asymptote: \(x = 2\) (\(f \to +\infty\) from left, \(f \to -\infty\) from right)
• Oblique asymptote: \(y = x\)
• Zeros: \(x = -1\) and \(x = 3\)
• \(y\)-intercept: \((0, \frac{3}{2})\)
• No critical points; strictly increasing on each branch
• Concave up for \(x < 2\); concave down for \(x > 2\)
• The curve approaches the asymptote \(y = x\) from below for \(x \to -\infty\) and from above for \(x \to +\infty\) (since \(-\frac{3}{x-2}\) is positive when \(x < 2\) and negative when \(x > 2\))

Problem 9: Funktionsscharen Parameter Analysis (xxx)

For \(f_t(x) = x^3 - 3t^2x + 2t^3\) with \(t > 0\):

1. Show that \(x = t\) is always a zero of \(f_t\).

2. Factor \(f_t(x)\) completely.

3. Find all extrema as a function of \(t\).

4. For which value of \(t\) does the local minimum have a \(y\)-value of \(-16\)?

CautionSolution

a) Substitute \(x = t\) into \(f_t(x)\):

\[f_t(t) = t^3 - 3t^2 \cdot t + 2t^3 = t^3 - 3t^3 + 2t^3 = 0 \checkmark\]

So \(x = t\) is always a zero of \(f_t\).

b) Since \((x - t)\) is a factor, perform polynomial division:

\[x^3 - 3t^2 x + 2t^3 = (x - t)(x^2 + tx - 2t^2)\]

Factor the quadratic: \(x^2 + tx - 2t^2 = (x + 2t)(x - t)\)

Therefore:

\[f_t(x) = (x - t)^2(x + 2t)\]

The zeros are \(x = t\) (double zero) and \(x = -2t\) (simple zero).

c) Find \(f_t'(x)\):

\[f_t'(x) = 3x^2 - 3t^2 = 3(x^2 - t^2) = 3(x-t)(x+t)\]

Critical points: \(x = t\) and \(x = -t\).

Find \(f_t''(x) = 6x\):

• At \(x = -t\) (with \(t > 0\)): \(f_t''(-t) = -6t < 0\) \(\Rightarrow\) local maximum
  • \(f_t(-t) = -t^3 + 3t^3 + 2t^3 = 4t^3\)
  • Local maximum at \((-t, 4t^3)\)
• At \(x = t\) (with \(t > 0\)): \(f_t''(t) = 6t > 0\) \(\Rightarrow\) local minimum
  • \(f_t(t) = 0\) (the double zero)
  • Local minimum at \((t, 0)\)

d) The local minimum value is \(f_t(t) = 0\) for all \(t > 0\).

This means the local minimum \(y\)-value is always \(0\), and it can never equal \(-16\).

However, let us reconsider. Perhaps we should also look at whether the question refers to a different interpretation. Notice that the local maximum value is \(4t^3\). If the question intended the local maximum to have a specific value:

\[4t^3 = -16 \implies t^3 = -4\]

This gives \(t = -\sqrt[3]{4}\), which violates \(t > 0\).

Re-examining: with \(f_t(x) = (x-t)^2(x+2t)\), the local minimum at \(x = t\) always gives \(y = 0\). So there is no value of \(t > 0\) for which the local minimum has \(y\)-value \(-16\).

Note: If we instead consider the family \(g_t(x) = x^3 - 3t^2x\) (without the \(+2t^3\)), then \(g_t(t) = t^3 - 3t^3 = -2t^3\), and setting \(-2t^3 = -16\) gives \(t^3 = 8\), so \(t = 2\).

With the given function \(f_t(x) = x^3 - 3t^2x + 2t^3\), the answer is: no such \(t\) exists (the local minimum value is always 0).

Problem 10: Curve Sketch of \(f(x) = x \cdot \ln(x)\) (xxx)

1. Determine the domain.

2. Find the zeros.

3. Compute \(\lim_{x \to 0^+} x \ln(x)\). (Hint: rewrite as \(\frac{\ln(x)}{1/x}\) and apply L’Hôpital’s rule.)

4. Find \(f'(x)\) and the critical points.

5. Find \(f''(x)\) and the inflection points.

6. Sketch the complete graph.

7. Find the minimum value and interpret it.

CautionSolution

a) The natural logarithm requires \(x > 0\), so:

\[D_f = (0, \infty)\]

b) Set \(f(x) = x \ln(x) = 0\).

Since \(x > 0\), we need \(\ln(x) = 0\), which gives \(x = 1\).

Zero at \(x = 1\): \(f(1) = 1 \cdot \ln(1) = 0\).

c) We evaluate \(\lim_{x \to 0^+} x \ln(x)\). This is of the indeterminate form \(0 \cdot (-\infty)\).

Rewrite:

\[\lim_{x \to 0^+} x \ln(x) = \lim_{x \to 0^+} \frac{\ln(x)}{1/x} = \lim_{x \to 0^+} \frac{\ln(x)}{x^{-1}}\]

This is of the form \(\frac{-\infty}{\infty}\), so we apply L’Hôpital’s rule:

\[= \lim_{x \to 0^+} \frac{1/x}{-x^{-2}} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \frac{x^2}{-x} = \lim_{x \to 0^+} (-x) = 0\]

So \(\lim_{x \to 0^+} x \ln(x) = 0\). The graph approaches the origin from below.

d) Apply the product rule:

\[f'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1\]

Set \(f'(x) = 0\):

\[\ln(x) + 1 = 0 \implies \ln(x) = -1 \implies x = e^{-1} = \frac{1}{e}\]

Critical point at \(x = \frac{1}{e}\).

e) Find \(f''(x)\):

\[f''(x) = \frac{1}{x}\]

Since \(x > 0\), we have \(f''(x) = \frac{1}{x} > 0\) for all \(x\) in the domain.

There are no inflection points — the function is concave up everywhere on \((0, \infty)\).

f) Key features for the sketch:

• Domain: \((0, \infty)\)
• Zero at \(x = 1\)
• \(\lim_{x \to 0^+} f(x) = 0\) (approaches origin)
• Minimum at \(x = \frac{1}{e} \approx 0.368\)
• No inflection points; concave up everywhere
• As \(x \to \infty\): \(f(x) \to \infty\)
• For \(0 < x < 1\): \(\ln(x) < 0\), so \(f(x) < 0\)
• For \(x > 1\): \(\ln(x) > 0\), so \(f(x) > 0\)

g) The minimum value occurs at \(x = \frac{1}{e}\):

\[f\left(\frac{1}{e}\right) = \frac{1}{e} \cdot \ln\left(\frac{1}{e}\right) = \frac{1}{e} \cdot (-1) = -\frac{1}{e} \approx -0.368\]

Interpretation: The function \(x \ln(x)\) attains its smallest value of \(-\frac{1}{e}\) at \(x = \frac{1}{e}\). This is a global minimum since the function is concave up everywhere on its domain.

Problem 11: Tangent Line System (xxx)

The tangent line to \(f(x) = ax^2 + bx + 1\) at \(x = 1\) has the equation \(y = 3x - 1\).

1. Set up a system of equations using \(f(1)\) and \(f'(1)\).

2. Solve for \(a\) and \(b\).

3. Sketch \(f(x)\) and the tangent line.

CautionSolution

a) The tangent line touches the curve at \(x = 1\), so:

Condition 1 — the point lies on the curve:

\[f(1) = a(1)^2 + b(1) + 1 = a + b + 1\]

The tangent line at \(x = 1\) gives \(y = 3(1) - 1 = 2\), so:

\[a + b + 1 = 2 \implies a + b = 1 \quad \text{...(I)}\]

Condition 2 — the slope matches:

\[f'(x) = 2ax + b \implies f'(1) = 2a + b\]

The slope of the tangent line \(y = 3x - 1\) is \(3\), so:

\[2a + b = 3 \quad \text{...(II)}\]

b) Subtract equation (I) from equation (II):

\[(2a + b) - (a + b) = 3 - 1 \implies a = 2\]

Substitute back into (I): \(2 + b = 1 \implies b = -1\)

Therefore \(a = 2\) and \(b = -1\), giving:

\[f(x) = 2x^2 - x + 1\]

Verification:

• \(f(1) = 2 - 1 + 1 = 2 = 3(1) - 1\) \(\checkmark\)
• \(f'(1) = 2(2)(1) - 1 = 3\) \(\checkmark\)

c) Key features for the sketch:

• \(f(x) = 2x^2 - x + 1\) is an upward-opening parabola
• Vertex: \(x = \frac{1}{4}\), \(f\left(\frac{1}{4}\right) = 2\left(\frac{1}{16}\right) - \frac{1}{4} + 1 = \frac{1}{8} - \frac{1}{4} + 1 = \frac{7}{8}\)
• The tangent line \(y = 3x - 1\) touches the parabola at the point \((1, 2)\)
• The tangent lies below the parabola everywhere except at the point of tangency (since the parabola is convex)

Problem 12: Cost Minimization with e-Function (xxx)

A company’s production cost per unit is modeled by:

\[C(x) = 50 + \frac{200}{x} + 0.1x \cdot e^{0.01x}\]

where \(x > 0\) is the number of units produced daily.

1. What is the cost per unit when \(x = 50\) units are produced?

2. Find \(C'(x)\).

3. Show that there exists a production level that minimizes the cost per unit. (Argue using the behavior as \(x \to 0^+\) and \(x \to \infty\).)

4. Use a numerical/graphical argument to estimate the optimal production level.

CautionSolution

a) Substitute \(x = 50\):

\[C(50) = 50 + \frac{200}{50} + 0.1 \cdot 50 \cdot e^{0.01 \cdot 50}\]

\[= 50 + 4 + 5 \cdot e^{0.5} = 54 + 5 \cdot 1.6487 \approx 54 + 8.24 = 62.24\]

The cost per unit at 50 units daily is approximately €62.24.

b) Differentiate term by term:

• \(\frac{d}{dx}(50) = 0\)
• \(\frac{d}{dx}\left(\frac{200}{x}\right) = -\frac{200}{x^2}\)
• For \(\frac{d}{dx}\left(0.1x \cdot e^{0.01x}\right)\), apply the product rule:

\[= 0.1 \cdot e^{0.01x} + 0.1x \cdot 0.01 \cdot e^{0.01x} = 0.1 e^{0.01x}(1 + 0.01x)\]

Therefore:

\[C'(x) = -\frac{200}{x^2} + 0.1 e^{0.01x}(1 + 0.01x)\]

c) Analyze the boundary behavior:

• As \(x \to 0^+\): The term \(\frac{200}{x} \to \infty\), so \(C(x) \to \infty\).
• As \(x \to \infty\): The term \(0.1x \cdot e^{0.01x} \to \infty\) (exponential growth dominates), so \(C(x) \to \infty\).

Since \(C(x) \to \infty\) at both ends of the domain and \(C\) is continuous on \((0, \infty)\), by the Extreme Value Theorem (applied to any closed interval containing all low values), \(C\) must attain a global minimum at some interior point. Therefore, there exists a production level that minimizes cost per unit.

d) We need to solve \(C'(x) = 0\):

\[\frac{200}{x^2} = 0.1 e^{0.01x}(1 + 0.01x)\]

This equation cannot be solved analytically. We evaluate \(C'(x)\) at selected points:

\(x\)	\(-\frac{200}{x^2}\)	\(0.1e^{0.01x}(1+0.01x)\)	\(C'(x)\)
10	\(-2.000\)	\(0.111\)	\(-1.889\)
20	\(-0.500\)	\(0.122\)	\(-0.378\)
30	\(-0.222\)	\(0.135\)	\(-0.087\)
35	\(-0.163\)	\(0.142\)	\(-0.021\)
40	\(-0.125\)	\(0.149\)	\(+0.024\)
37	\(-0.146\)	\(0.145\)	\(-0.001\)
38	\(-0.138\)	\(0.147\)	\(+0.009\)

The sign change occurs between \(x = 37\) and \(x = 38\), so the optimal production level is approximately \(x \approx 37\) units per day.

At this level: \(C(37) \approx 50 + \frac{200}{37} + 0.1 \cdot 37 \cdot e^{0.37} \approx 50 + 5.41 + 3.7 \cdot 1.448 \approx 50 + 5.41 + 5.36 \approx 60.77\).

Interpretation: The company should produce approximately 37 units per day to minimize the cost per unit at roughly €60.77.

Problem 13: Oblique Asymptote + Full Sketch (xxx)

For \(f(x) = \frac{2x^2 - x - 6}{x + 2}\):

1. Perform polynomial long division.

2. Identify the oblique and vertical asymptotes.

3. Find all zeros.

4. Find \(f'(x)\) and the critical points.

5. Sketch the graph with the oblique asymptote.

CautionSolution

a) Divide \(2x^2 - x - 6\) by \(x + 2\):

• \(2x^2 \div x = 2x\). Multiply: \(2x(x+2) = 2x^2 + 4x\). Subtract: \((2x^2 - x - 6) - (2x^2 + 4x) = -5x - 6\).
• \(-5x \div x = -5\). Multiply: \(-5(x+2) = -5x - 10\). Subtract: \((-5x - 6) - (-5x - 10) = 4\).

\[f(x) = 2x - 5 + \frac{4}{x+2}\]

b) Asymptotes:

• Oblique asymptote: \(y = 2x - 5\) (as \(x \to \pm \infty\), the remainder \(\frac{4}{x+2} \to 0\))
• Vertical asymptote: \(x = -2\) (denominator is zero, numerator is \(2(4) - (-2) - 6 = 4 \neq 0\) at \(x = -2\)… let’s check: \(2(-2)^2 - (-2) - 6 = 8 + 2 - 6 = 4 \neq 0\), confirmed)

Behavior near \(x = -2\):

• As \(x \to -2^+\): \(\frac{4}{x+2} \to +\infty\), so \(f(x) \to +\infty\)
• As \(x \to -2^-\): \(\frac{4}{x+2} \to -\infty\), so \(f(x) \to -\infty\)

c) Set numerator to zero: \(2x^2 - x - 6 = 0\).

Using the quadratic formula:

\[x = \frac{1 \pm \sqrt{1 + 48}}{4} = \frac{1 \pm 7}{4}\]

\[x = 2 \quad \text{or} \quad x = -\frac{3}{2}\]

Zeros at \(x = 2\) and \(x = -\frac{3}{2}\).

\(y\)-intercept: \(f(0) = \frac{-6}{2} = -3\).

d) Using \(f(x) = 2x - 5 + 4(x+2)^{-1}\):

\[f'(x) = 2 - \frac{4}{(x+2)^2}\]

Set \(f'(x) = 0\):

\[2 = \frac{4}{(x+2)^2} \implies (x+2)^2 = 2 \implies x + 2 = \pm\sqrt{2}\]

\[x = -2 + \sqrt{2} \approx -0.586 \quad \text{or} \quad x = -2 - \sqrt{2} \approx -3.414\]

Classify using \(f''(x) = \frac{8}{(x+2)^3}\):

• At \(x = -2 + \sqrt{2}\): \((x+2)^3 = (\sqrt{2})^3 > 0\), so \(f''> 0\) \(\Rightarrow\) local minimum
  • \(f(-2+\sqrt{2}) = 2(-2+\sqrt{2}) - 5 + \frac{4}{\sqrt{2}} = -4 + 2\sqrt{2} - 5 + 2\sqrt{2} = -9 + 4\sqrt{2} \approx -3.34\)
• At \(x = -2 - \sqrt{2}\): \((x+2)^3 = (-\sqrt{2})^3 < 0\), so \(f'' < 0\) \(\Rightarrow\) local maximum
  • \(f(-2-\sqrt{2}) = 2(-2-\sqrt{2}) - 5 + \frac{4}{-\sqrt{2}} = -4 - 2\sqrt{2} - 5 - 2\sqrt{2} = -9 - 4\sqrt{2} \approx -14.66\)

e) Key features for the sketch:

• Vertical asymptote: \(x = -2\)
• Oblique asymptote: \(y = 2x - 5\)
• Zeros: \(x = -1.5\) and \(x = 2\)
• \(y\)-intercept: \((0, -3)\)
• Local minimum at \((-2 + \sqrt{2}, -9 + 4\sqrt{2}) \approx (-0.59, -3.34)\) — right branch
• Local maximum at \((-2 - \sqrt{2}, -9 - 4\sqrt{2}) \approx (-3.41, -14.66)\) — left branch
• Right branch: curve approaches asymptote from above (since \(\frac{4}{x+2} > 0\) for \(x > -2\))
• Left branch: curve approaches asymptote from below (since \(\frac{4}{x+2} < 0\) for \(x < -2\))

Problem 14: Related Rates Word Problem (xxx)

A spherical balloon is inflated so that its volume increases at a constant rate of \(100 \text{ cm}^3/\text{s}\).

1. Express the volume as a function of the radius: \(V = \frac{4}{3}\pi r^3\).

2. Find \(\frac{dr}{dt}\) when \(r = 10\) cm.

3. Find \(\frac{dA}{dt}\) when \(r = 10\) cm, where \(A = 4\pi r^2\) is the surface area.

4. At what radius is the surface area increasing at a rate of \(40 \text{ cm}^2/\text{s}\)?

CautionSolution

a) The volume of a sphere with radius \(r\) is:

\[V = \frac{4}{3}\pi r^3\]

We are given that \(\frac{dV}{dt} = 100 \text{ cm}^3/\text{s}\).

b) Differentiate \(V\) with respect to \(t\) using the chain rule:

\[\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}\]

Solve for \(\frac{dr}{dt}\):

\[\frac{dr}{dt} = \frac{1}{4\pi r^2} \cdot \frac{dV}{dt}\]

At \(r = 10\) cm:

\[\frac{dr}{dt} = \frac{100}{4\pi (10)^2} = \frac{100}{400\pi} = \frac{1}{4\pi} \approx 0.0796 \text{ cm/s}\]

c) The surface area is \(A = 4\pi r^2\). Differentiate with respect to \(t\):

\[\frac{dA}{dt} = 8\pi r \cdot \frac{dr}{dt}\]

At \(r = 10\) cm, using \(\frac{dr}{dt} = \frac{1}{4\pi}\):

\[\frac{dA}{dt} = 8\pi (10) \cdot \frac{1}{4\pi} = \frac{80\pi}{4\pi} = 20 \text{ cm}^2/\text{s}\]

Alternatively, we can derive a direct relationship. From \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\) and \(\frac{dA}{dt} = 8\pi r \frac{dr}{dt}\):

\[\frac{dA}{dt} = \frac{8\pi r}{4\pi r^2} \cdot \frac{dV}{dt} = \frac{2}{r} \cdot \frac{dV}{dt}\]

At \(r = 10\): \(\frac{dA}{dt} = \frac{2}{10} \cdot 100 = 20 \text{ cm}^2/\text{s}\) \(\checkmark\)

d) Using \(\frac{dA}{dt} = \frac{2}{r} \cdot \frac{dV}{dt}\):

\[40 = \frac{2}{r} \cdot 100 = \frac{200}{r}\]

\[r = \frac{200}{40} = 5 \text{ cm}\]

Interpretation: When the balloon has a radius of 5 cm, its surface area is increasing at a rate of 40 cm²/s. Notice that as the balloon grows larger, the surface area increases more slowly (for a constant rate of volume increase), because the same volume of air is spread over a larger surface.

Problem 15: Advertising Effectiveness Full Analysis (xxxx)

An advertising campaign’s effectiveness is modeled by:

\[E(t) = 500(1 - e^{-0.3t})\]

where \(E(t)\) is in thousands of impressions reached and \(t\) is weeks since the campaign started.

1. Find \(E(0)\) and interpret the result.

2. Find \(\lim_{t \to \infty} E(t)\) and interpret.

3. Find \(E'(t)\) and interpret \(E'(0)\).

4. Find \(E''(t)\). Is the campaign’s effectiveness concave up or concave down? Interpret this in the context of advertising.

5. The campaign costs €2,000 per week. Revenue per thousand impressions is €8. Find the time \(t^*\) at which marginal revenue equals marginal cost.

6. Calculate the total profit at \(t^*\).

CautionSolution

a) Evaluate at \(t = 0\):

\[E(0) = 500(1 - e^{0}) = 500(1 - 1) = 0\]

Interpretation: At the start of the campaign (\(t = 0\)), no impressions have been reached yet. The campaign has just begun.

b) Compute the limit:

\[\lim_{t \to \infty} E(t) = 500 \lim_{t \to \infty}(1 - e^{-0.3t}) = 500(1 - 0) = 500\]

Interpretation: The campaign can reach a maximum of 500,000 impressions (500 thousand). This is the saturation level — no matter how long the campaign runs, it cannot exceed this ceiling. This reflects the finite size of the target audience.

c) Differentiate:

\[E'(t) = 500 \cdot 0.3 \cdot e^{-0.3t} = 150 e^{-0.3t}\]

At \(t = 0\):

\[E'(0) = 150 e^{0} = 150\]

Interpretation: At the start of the campaign, the reach is growing at a rate of 150,000 impressions per week. This is the fastest rate of growth — the campaign is most effective in its earliest phase.

d) Differentiate again:

\[E''(t) = 150 \cdot (-0.3) \cdot e^{-0.3t} = -45 e^{-0.3t}\]

Since \(e^{-0.3t} > 0\) for all \(t\), we have \(E''(t) < 0\) for all \(t \geq 0\).

The function is concave down everywhere.

Interpretation: The rate at which new impressions are gained is always decreasing. Each additional week of advertising reaches fewer new people than the previous week. This is the law of diminishing returns in advertising — the easiest-to-reach audience is captured first, and later weeks yield progressively smaller gains.

e) The revenue function is:

\[R(t) = 8 \cdot E(t) = 8 \cdot 500(1 - e^{-0.3t}) = 4000(1 - e^{-0.3t}) \quad \text{(in €)}\]

Marginal revenue (revenue gained per additional week):

\[R'(t) = 4000 \cdot 0.3 \cdot e^{-0.3t} = 1200 e^{-0.3t}\]

The cost is €2,000 per week, so marginal cost is constant:

\[MC = 2000\]

Set marginal revenue equal to marginal cost:

\[1200 e^{-0.3t^*} = 2000\]

\[e^{-0.3t^*} = \frac{2000}{1200} = \frac{5}{3}\]

Since \(\frac{5}{3} > 1\) and \(e^{-0.3t} \leq 1\) for \(t \geq 0\), this equation has no solution for \(t \geq 0\).

This means \(R'(0) = 1200 < 2000 = MC\): the marginal revenue is always less than the marginal cost. The campaign’s marginal revenue at its very best (week 0) is only €1,200, which is already below the weekly cost of €2,000.

Therefore \(t^* = 0\): it is never profitable to run this campaign at these cost parameters. The optimal decision is not to run the campaign at all.

f) Since \(t^* = 0\), the total profit is:

\[\Pi(0) = R(0) - C(0) = 0 - 0 = 0\]

If the campaign were run for \(t\) weeks anyway, the profit would be:

\[\Pi(t) = R(t) - 2000t = 4000(1 - e^{-0.3t}) - 2000t\]

We can verify: \(\Pi'(t) = 1200 e^{-0.3t} - 2000 < 0\) for all \(t \geq 0\) (since \(1200 < 2000\)). So profit is always decreasing — every week of the campaign loses money.

Interpretation: At a cost of €2,000 per week and revenue of €8 per thousand impressions, this advertising campaign is not economically viable. The maximum marginal revenue of €1,200/week (at the start) never covers the weekly cost. The company should either negotiate a lower weekly cost, find a way to increase revenue per impression, or abandon this campaign.

Problem 16: Tangent from External Point (xxxx)

Find all tangent lines from the point \(P(0, -4)\) to the curve \(f(x) = x^3 - 3x\).

1. Let the tangent touch the curve at \((a, f(a))\). Write the tangent line equation.

2. Use the condition that the tangent passes through \(P(0, -4)\) to obtain an equation in \(a\).

3. Solve for \(a\).

4. Write the equations of all tangent lines.

5. Sketch the curve and all tangent lines.

CautionSolution

a) At the point \((a, f(a)) = (a, a^3 - 3a)\), the slope of the tangent is:

\[f'(a) = 3a^2 - 3\]

The tangent line equation (point-slope form):

\[y - (a^3 - 3a) = (3a^2 - 3)(x - a)\]

Expanding:

\[y = (3a^2 - 3)x - 3a^3 + 3a + a^3 - 3a = (3a^2 - 3)x - 2a^3\]

So the tangent line at \(x = a\) is:

\[y = (3a^2 - 3)x - 2a^3\]

b) The tangent must pass through \(P(0, -4)\). Substitute \(x = 0\), \(y = -4\):

\[-4 = (3a^2 - 3)(0) - 2a^3\]

\[-4 = -2a^3\]

\[a^3 = 2\]

c) Solve \(a^3 = 2\):

\[a = \sqrt[3]{2} \approx 1.26\]

There is exactly one real solution (since \(x^3 = 2\) has one real root and two complex roots).

d) With \(a = \sqrt[3]{2}\):

The slope is:

\[m = 3a^2 - 3 = 3(\sqrt[3]{2})^2 - 3 = 3\sqrt[3]{4} - 3 = 3(\sqrt[3]{4} - 1) \approx 3(1.587 - 1) = 1.762\]

The tangent line:

\[y = (3\sqrt[3]{4} - 3)x - 2(\sqrt[3]{2})^3 = (3\sqrt[3]{4} - 3)x - 4\]

In decimal form:

\[y \approx 1.762x - 4\]

Verification: At \(x = 0\): \(y = -4\) \(\checkmark\)

At \(x = a = \sqrt[3]{2}\): \(y = (3\sqrt[3]{4} - 3)\sqrt[3]{2} - 4 = 3\sqrt[3]{8} - 3\sqrt[3]{2} - 4 = 6 - 3\sqrt[3]{2} - 4 = 2 - 3\sqrt[3]{2}\)

Check \(f(a) = (\sqrt[3]{2})^3 - 3\sqrt[3]{2} = 2 - 3\sqrt[3]{2}\) \(\checkmark\)

e) Key features for the sketch:

• \(f(x) = x^3 - 3x\) has zeros at \(x = 0, \pm\sqrt{3}\)
• Local maximum at \(x = -1\): \(f(-1) = 2\)
• Local minimum at \(x = 1\): \(f(1) = -2\)
• Inflection point at \(x = 0\): \(f(0) = 0\)
• The point \(P(0, -4)\) lies below the curve
• One tangent line from \(P\) touches the curve at \((\sqrt[3]{2}, 2 - 3\sqrt[3]{2}) \approx (1.26, -1.78)\)
• The tangent line passes through \((0, -4)\) with slope \(\approx 1.76\)

Problem 17: Funktionsscharen with Area Constraint (xxxx)

For \(f_a(x) = ax^2 - 2ax\) with \(a > 0\):

1. Find the zeros and the vertex as functions of \(a\).

2. Sketch \(f_1(x)\) and \(f_2(x)\).

3. The region between \(f_a(x)\) and the \(x\)-axis has area \(\frac{4a}{3}\). Verify this by computing the definite integral.

4. Find \(a\) such that the enclosed area equals 12.

5. Find the tangent line to \(f_a\) at \(x = 0\) as a function of \(a\). For which value of \(a\) does this tangent line pass through the point \((3, -12)\)?

CautionSolution

a) Factor: \(f_a(x) = ax(x - 2)\)

Zeros: \(x = 0\) and \(x = 2\) (independent of \(a\)!)

Vertex: The vertex lies at \(x = \frac{0 + 2}{2} = 1\) (midpoint of the zeros).

\[f_a(1) = a(1)^2 - 2a(1) = a - 2a = -a\]

The vertex is at \((1, -a)\).

Note: Since \(a > 0\), the parabola opens upward and the vertex lies below the \(x\)-axis.

b) For \(f_1(x) = x^2 - 2x\):

• Zeros: \(x = 0, 2\)
• Vertex: \((1, -1)\)

For \(f_2(x) = 2x^2 - 4x\):

• Zeros: \(x = 0, 2\)
• Vertex: \((1, -2)\)

Both parabolas share the same zeros but \(f_2\) is narrower (steeper) with a deeper vertex.

c) The curve lies below the \(x\)-axis on \([0, 2]\) (since \(a > 0\) and the parabola opens upward). The enclosed area is:

\[A = \int_0^2 |f_a(x)| \, dx = -\int_0^2 f_a(x) \, dx = -\int_0^2 (ax^2 - 2ax) \, dx\]

Compute the integral:

\[-\int_0^2 (ax^2 - 2ax) \, dx = -\left[\frac{ax^3}{3} - ax^2\right]_0^2\]

\[= -\left(\frac{8a}{3} - 4a\right) = -\left(\frac{8a - 12a}{3}\right) = -\left(\frac{-4a}{3}\right) = \frac{4a}{3}\]

This confirms that the enclosed area is \(\frac{4a}{3}\). \(\checkmark\)

d) Set the area equal to 12:

\[\frac{4a}{3} = 12 \implies 4a = 36 \implies a = 9\]

Verification: \(f_9(x) = 9x^2 - 18x\). Area \(= \frac{4 \cdot 9}{3} = 12\) \(\checkmark\)

e) The tangent line at \(x = 0\):

\[f_a'(x) = 2ax - 2a \implies f_a'(0) = -2a\]

The point of tangency is \((0, f_a(0)) = (0, 0)\).

The tangent line is:

\[y = -2a \cdot x\]

For this tangent to pass through \((3, -12)\):

\[-12 = -2a \cdot 3 = -6a\]

\[a = 2\]

Verification: With \(a = 2\), the tangent line is \(y = -4x\). At \(x = 3\): \(y = -12\) \(\checkmark\)

Also verify that this is indeed tangent to \(f_2(x) = 2x^2 - 4x\) at \(x = 0\):

• \(f_2(0) = 0\) and the line gives \(y(0) = 0\) \(\checkmark\)
• \(f_2'(0) = -4\) and the slope of the line is \(-4\) \(\checkmark\)