Tasks 03 - Functions

Mathematics for Master Students

These problems accompany Session 03 of the tutorial. Work through them at your own pace before the next session. We start Session 04 with your questions. On the website, every problem has a fully worked solution you can expand to check yourself. The PDF version contains only the problems, so you can print it and use it as a worksheet.

The problems are staged: the first two are warm-ups, the middle ones practice the core techniques, and the last two are at the level of the mock exam.

Problem 1: Evaluating a Function

Let \(f(x) = x^2 + 4x\).

Compute \(f(3)\), \(f(0)\) and \(f(-2)\).
Compute \(f\left(\frac{1}{2}\right)\).
Compute \(f(a + h)\) and expand the result.
Check with \(a = h = 1\) that \(f(a + h) \neq f(a) + f(h)\).

Solution

Evaluating means substituting: replace \(x\) by the input everywhere it appears, negative inputs in parentheses.

Substitute one input at a time: \[f(3) = 3^2 + 4 \cdot 3 = 9 + 12 = 21\] \[f(0) = 0^2 + 4 \cdot 0 = 0\] \[f(-2) = (-2)^2 + 4 \cdot (-2) = 4 - 8 = -4\] Note the parentheses in the last line: \((-2)^2 = 4\), not \(-4\).
Fractions work exactly the same way: \[f\left(\tfrac{1}{2}\right) = \left(\tfrac{1}{2}\right)^2 + 4 \cdot \tfrac{1}{2} = \tfrac{1}{4} + 2 = \tfrac{9}{4}\]
Substitute the whole expression \(a + h\), in parentheses, then expand with the binomial formula: \[f(a + h) = (a + h)^2 + 4(a + h) = a^2 + 2ah + h^2 + 4a + 4h\]
With \(a = h = 1\): the left side is \(f(1 + 1) = f(2) = 4 + 8 = 12\), while the right side is \[f(1) + f(1) = (1 + 4) + (1 + 4) = 10\] Since \(12 \neq 10\), we confirm \(f(a + h) \neq f(a) + f(h)\): a function is not a number you can “multiply out” of a bracket. Expressions like \(f(a + h)\) return in Session 04, where they measure how much \(f\) changes when the input moves by \(h\).

Problem 2: Domain, Codomain, Range

A parcel service charges by weight class. For the classes \(A = \{1, 2, 3, 4, 5\}\) (kilograms, rounded up), the fee in € is given by

\[f: A \to \mathbb{R}, \quad f(x) = 4x + 6\]

State the domain and the codomain of \(f\), and compute its range.
Now consider \(g: \mathbb{R} \to \mathbb{R}\), \(g(x) = x^2 + 2\). What is the range of \(g\), and why is it not equal to the codomain?
The service also keeps a data table assigning to each customer the order values of all their orders. One customer placed three orders. Is “customer \(\mapsto\) order value” a function? What about “order \(\mapsto\) order value”?
True or false: the range of a function is always a subset of its codomain.

Solution

The domain is the set of allowed inputs, \(A = \{1, 2, 3, 4, 5\}\); the codomain is where outputs are declared to land, \(\mathbb{R}\). The range collects the outputs that actually occur: evaluate \(f\) on every element of the domain: \[f(1) = 10, \quad f(2) = 14, \quad f(3) = 18, \quad f(4) = 22, \quad f(5) = 26\] So the range is \(f(A) = \{10, 14, 18, 22, 26\}\), a finite set of five fees.
Squares are never negative, so \(x^2 + 2 \geq 2\) for every \(x\), and every value \(y \geq 2\) actually occurs (e.g. \(g(\sqrt{y - 2}) = y\)). The range is therefore \([2, \infty)\). The codomain \(\mathbb{R}\) is just the declared target: no output is ever negative, so the range is a strictly smaller set. That is perfectly fine: the definition only requires outputs to land inside the codomain.
“Customer \(\mapsto\) order value” is not a function: the customer with three orders is one input with three different outputs, violating “exactly one output per input”. “Order \(\mapsto\) order value” is a function: every single order has exactly one value. Choosing the right input set is what turns messy data into a function.
True. By definition every output \(f(x)\) lands in the codomain \(B\), so \(f(A) \subseteq B\) always holds. Equality is possible but not required, as b) shows.

Problem 3: Maximal Domains

If no domain is stated, we use the maximal domain: the largest set of real numbers on which the formula is defined. Find it for each function and write it in interval notation.

\(f(x) = \dfrac{7}{2x - 10}\)
\(g(x) = \sqrt{9 - 3x}\)
\(h(x) = \ln(2x + 8)\)
\(k(x) = \dfrac{\sqrt{x + 1}}{x - 2}\)

Solution

Work through the three red flags: no division by zero, no even roots of negatives, no logs of non-positives.

Denominator \(\neq 0\): we need \(2x - 10 \neq 0\), i.e. \(x \neq 5\). Everything else is fine, so the maximal domain is \[\mathbb{R} \setminus \{5\} = (-\infty, 5) \cup (5, \infty)\]
Root needs \(\geq 0\): we need \(9 - 3x \geq 0\), i.e. \(9 \geq 3x\), i.e. \(x \leq 3\). The maximal domain is \[(-\infty, 3]\] Square bracket at \(3\): the value \(g(3) = \sqrt{0} = 0\) is defined.
Log needs \(> 0\), strictly: we need \(2x + 8 > 0\), i.e. \(x > -4\). The maximal domain is \[(-4, \infty)\] Round bracket at \(-4\): \(\ln 0\) does not exist, so the endpoint is excluded, unlike the root in b).
Two flags at once, and both conditions must hold (an intersection, as in Session 02):

Step 1 - The root: \(x + 1 \geq 0\), so \(x \geq -1\).

Step 2 - The denominator: \(x - 2 \neq 0\), so \(x \neq 2\).

Step 3 - Combine: start from \([-1, \infty)\) and remove the point \(2\): \[[-1, 2) \cup (2, \infty)\]

Common Mistake

Finding one restriction and stopping. Collect every restriction first, then intersect them, and remember the log is strict: \(\ln\) needs \(> 0\), while an even root is happy with \(\geq 0\).

Problem 4: Break-Even Analysis

A manufacturer of e-bike batteries has fixed costs of €30,000 per month. Each battery costs €75 to produce and sells for €125. Let \(x\) be the number of batteries produced and sold per month.

Write down the monthly cost function \(C(x)\) and the revenue function \(R(x)\). What do the slope and the intercept of \(C\) mean in business terms?
Compute cost, revenue and the resulting profit or loss at \(x = 400\) units.
Find the break-even point: the quantity at which revenue exactly covers cost. Verify your answer.
Write down the profit function \(P(x) = R(x) - C(x)\). How many batteries must be sold to reach a monthly profit of €20,000?

Solution

Both functions are linear: \[C(x) = 30{,}000 + 75x \qquad R(x) = 125x\] In \(C\), the intercept \(30{,}000 = C(0)\) is the fixed cost: it is due even at zero production. The slope \(75\) is the variable cost per unit: each extra battery adds €75 of cost. \(R\) has no fixed part: no sales, no revenue.
Substitute \(x = 400\): \[C(400) = 30{,}000 + 75 \cdot 400 = 30{,}000 + 30{,}000 = 60{,}000\] \[R(400) = 125 \cdot 400 = 50{,}000\] Revenue €50,000 against cost €60,000, a loss of €10,000. At 400 units the fixed costs are not yet covered.
Break-even means \(C(x) = R(x)\): set the two functions equal and solve:

Step 1 - Set up the equation: \[30{,}000 + 75x = 125x\]

Step 2 - Collect the \(x\)-terms: \[30{,}000 = 125x - 75x = 50x\]

Step 3 - Solve: \[x = \frac{30{,}000}{50} = 600\]

Check: \(C(600) = 30{,}000 + 45{,}000 = 75{,}000\) and \(R(600) = 125 \cdot 600 = 75{,}000\). ✓ Below 600 units the month ends in a loss (as at \(x = 400\) in b), above 600 in a profit.
Subtracting the linear functions: \[P(x) = 125x - (30{,}000 + 75x) = 50x - 30{,}000\] The slope \(50 = 125 - 75\) is the contribution margin: every battery contributes €50 toward fixed costs and profit. For a profit of €20,000: \[50x - 30{,}000 = 20{,}000 \quad \Rightarrow \quad 50x = 50{,}000 \quad \Rightarrow \quad x = 1000\] Note that \(P(600) = 0\): the break-even point is exactly the root of the profit function.

Problem 5: Maximal Revenue and the Vertex

A regional carrier estimates the weekly demand for a new express connection as \(q(p) = 600 - 15p\), where \(p\) is the price in € and \(q\) the number of booked shipments.

Write down the revenue function \(R(p) = p \cdot q(p)\). What type of function is it, and which way does its parabola open?
Find the revenue-maximal price with the vertex formula \(p = -\frac{b}{2a}\), and compute the maximal revenue.
Confirm your result by completing the square: bring \(R(p)\) into the vertex form \(R(p) = a(p - p_0)^2 + R_0\) and read off the vertex. Why does this form prove that the vertex is a maximum?
For which prices is the model meaningful at all? Give the interval and interpret its endpoints.

Solution

Multiply out: \[R(p) = p \cdot (600 - 15p) = 600p - 15p^2\] A quadratic function with \(a = -15\), \(b = 600\), \(c = 0\). Since \(a < 0\), the parabola opens downward, the classic price trade-off: raising the price first raises revenue, then demand collapses.
The vertex formula gives \[p = -\frac{b}{2a} = -\frac{600}{2 \cdot (-15)} = -\frac{600}{-30} = 20\] Maximal revenue: \[R(20) = 600 \cdot 20 - 15 \cdot 20^2 = 12{,}000 - 6000 = 6000\] The best price is €20, yielding €6000 of weekly revenue (with \(q(20) = 300\) shipments).
Completing the square, step by step:

Step 1 - Factor out \(a = -15\) from the terms containing \(p\): \[R(p) = -15p^2 + 600p = -15\left(p^2 - 40p\right)\]

Step 2 - Complete the square inside: half of \(40\) is \(20\), and \(20^2 = 400\); add and subtract it: \[-15\left(p^2 - 40p + 400 - 400\right) = -15\left((p - 20)^2 - 400\right)\]

Step 3 - Multiply back out: \[R(p) = -15(p - 20)^2 + 6000\]

The vertex \((20, 6000)\) matches b). The vertex form proves the maximum: \((p - 20)^2 \geq 0\) for every \(p\), so \(-15(p - 20)^2 \leq 0\), and therefore \(R(p) \leq 6000\), with equality exactly at \(p = 20\). The formula \(p = -\frac{b}{2a}\) from the lecture is nothing but this computation done once and for all in general letters.
Two business restrictions: the price satisfies \(p \geq 0\), and demand must not be negative: \(600 - 15p \geq 0\), i.e. \(p \leq 40\). The meaningful domain is \[[0, 40]\] At \(p = 0\) we give the service away (\(R = 0\)); at \(p = 40\) demand, and with it revenue, hits zero. The revenue parabola is zero at both endpoints and peaks exactly in the middle, at \(p = 20\).

Problem 6: Composition and Inverses

An online shop applies a 20% discount, \(d(p) = 0.8p\), and adds a flat €5 shipping fee, \(s(p) = p + 5\), to a basket value \(p\).

Compute both compositions \((s \circ d)(p)\) and \((d \circ s)(p)\). Which order is cheaper for the customer, and by how much?
Evaluate both compositions for a basket of \(p = 60\).
The shop’s demand forecast is \(D(p) = 800 - 25p\). Find the inverse function \(D^{-1}\) with the swap-and-solve recipe, interpret it, and verify it with the pair \(p = 20\).
Why does \(h(x) = x^2\) on \(\mathbb{R}\) have no inverse, and how can the domain be repaired so that it does?

Solution

Inside out, the inner function acts first:

Discount first, then shipping: \[(s \circ d)(p) = s(0.8p) = 0.8p + 5\]

Shipping first, then discount: \[(d \circ s)(p) = d(p + 5) = 0.8(p + 5) = 0.8p + 4\]

The results differ: composition is not commutative. For the customer, \((d \circ s)\) is cheaper by exactly €1 on every order: when the discount is applied last, the €5 shipping fee gets discounted too (\(0.8 \cdot 5 = 4\)).
Substitute \(p = 60\): \[(s \circ d)(60) = 0.8 \cdot 60 + 5 = 48 + 5 = 53 \qquad (d \circ s)(60) = 0.8 \cdot 60 + 4 = 52\] A €60 basket costs €53 or €52, the promised €1 difference.
Swap-and-solve: write \(q = D(p)\) and solve for \(p\):

Step 1 - Set up: \(q = 800 - 25p\)

Step 2 - Isolate the term with \(p\): \(25p = 800 - q\)

Step 3 - Divide: \(p = \dfrac{800 - q}{25} = 32 - \dfrac{q}{25}\)

So \(D^{-1}(q) = 32 - \frac{q}{25}\): it answers the reverse question: which price must be set to sell exactly \(q\) units?

Check with a pair: \(D(20) = 800 - 500 = 300\), and \(D^{-1}(300) = 32 - \frac{300}{25} = 32 - 12 = 20\). ✓ Domain and range swap: meaningful quantities \(q \in [0, 800]\) map back to prices \(p \in [0, 32]\).
An inverse exists only if the function is one-to-one: every output must come from exactly one input. \(h(x) = x^2\) fails on \(\mathbb{R}\): \(h(3) = h(-3) = 9\), so “which input produced \(9\)?” has no unique answer: a horizontal line at height \(9\) hits the graph twice. Restricting the domain to \([0, \infty)\) repairs it: there \(h\) is strictly increasing, hence one-to-one, and \(h^{-1}(x) = \sqrt{x}\).

Common Mistake

\(D^{-1}\) is not \(\frac{1}{D}\): here \(D^{-1}(q) = 32 - \frac{q}{25}\), while \(\frac{1}{D(p)} = \frac{1}{800 - 25p}\), completely different objects. For functions, the exponent \(-1\) means undo, not divide.

Problem 7: Exponential Depreciation

A logistics company buys a forklift for €80,000. Its resale value depreciates by 15% per year, so after \(t\) years it is worth

\[V(t) = 80{,}000 \cdot 0.85^t\]

Explain the two constants in the formula. Then show, using an exponent rule, that \(V(t + 1) = 0.85 \cdot V(t)\) for every \(t\): each year costs the same percentage, not the same amount.
Compute the value after 3 years and after 5 years.
The company plans to replace the forklift once its value drops below €20,000. After how many years does that happen? Solve exactly with logarithms, then give the first full year in which the value is below the threshold.
Find the “half-value time”: how long until the forklift is worth half its purchase price? Use your result to double-check c).

Solution

\(80{,}000 = V(0)\) is the purchase price, and the base \(0.85 = 1 - 0.15\) is the factor that survives each year after losing 15%. By the exponent rule \(a^{x+y} = a^x a^y\): \[V(t + 1) = 80{,}000 \cdot 0.85^{t+1} = 80{,}000 \cdot 0.85^t \cdot 0.85 = 0.85 \cdot V(t)\] Every year multiplies the current value by \(0.85\), so the loss in € shrinks over time, even though the percentage stays fixed.
Substitute and evaluate the powers: \[V(3) = 80{,}000 \cdot 0.85^3 = 80{,}000 \cdot 0.614125 = 49{,}130\] \[V(5) = 80{,}000 \cdot 0.85^5 = 80{,}000 \cdot 0.44370\ldots \approx 35{,}496\] After 3 years the forklift is worth €49,130, after 5 years about €35,496, already less than half.
Set up the inequality and solve with logs:

Step 1 - Isolate the power: \[80{,}000 \cdot 0.85^t < 20{,}000 \quad \Rightarrow \quad 0.85^t < \frac{20{,}000}{80{,}000} = 0.25\]

Step 2 - Take \(\ln\) and pull down the exponent with the power rule \(\ln(x^k) = k \ln x\): \[t \cdot \ln 0.85 < \ln 0.25\]

Step 3 - Divide by \(\ln 0.85\). Careful: \(0.85 < 1\) means \(\ln 0.85 \approx -0.1625\) is negative, so the inequality flips: \[t > \frac{\ln 0.25}{\ln 0.85} \approx \frac{-1.3863}{-0.1625} \approx 8.53\]

The value crosses the threshold during the ninth year. First full year below €20,000: \(t = 9\), since \(V(8) \approx 80{,}000 \cdot 0.2725 \approx\) €21,799 is still above and \(V(9) \approx 80{,}000 \cdot 0.2316 \approx\) €18,529 is below.
Half the purchase price means \(0.85^t = 0.5\): \[t = \frac{\ln 0.5}{\ln 0.85} \approx \frac{-0.6931}{-0.1625} \approx 4.27 \text{ years}\] This is the decay twin of the lecture’s doubling time \(t = \frac{\ln 2}{r}\): the same log technique, base \(0.85\) instead of \(e^r\). (Equivalently, \(t = \log_{0.85} 0.5 = \frac{\ln 0.5}{\ln 0.85}\) by the change-of-base formula.)

Double-check of c): dropping to €20,000 means dropping to a quarter of the purchase price, and by the exponent rule \((a^x)^y = a^{xy}\), halving twice is one quarter: \[0.85^{2 \cdot 4.265} = \left(0.85^{4.265}\right)^2 \approx 0.5^2 = 0.25\] So the threshold time must be twice the half-value time: \(2 \cdot 4.265 \approx 8.53\). ✓

Problem 8: Reading Function Behavior

A planning team works with five model functions:

\[T(x) = 40\sqrt{x} \quad (x \geq 0) \qquad u(x) = \frac{600}{x} + 8 \quad (x > 0) \qquad V(t) = 20{,}000 \cdot 0.9^t \quad (t \geq 0)\]

\[L(x) = 120 - 4x \qquad q(x) = (x - 6)^2 + 2\]

(\(T\): daily throughput for \(x\) workers; \(u\): cost per unit at batch size \(x\); \(V\): resale value; \(L\): stock level after \(x\) weeks; \(q\): unit cost at machine speed \(x\).)

Classify each function as (strictly) increasing, (strictly) decreasing, or neither, with a one-line reason; no derivatives needed.
Classify each function as convex, concave, or both, using the chord test from the definition (chord above the graph = convex, below = concave).
Which of the five functions are one-to-one on the given domain, and therefore invertible?
Give a lower bound for \(u\), \(V\) and \(q\). Is \(T\) bounded above?

Solution

Reasoning input by input:
- \(T\) is strictly increasing: a larger \(x\) has a larger square root: more workers, more throughput.
- \(u\) is strictly decreasing on \(x > 0\): a larger batch divides the fixed €600 among more units, so \(\frac{600}{x}\) shrinks: economies of scale.
- \(V\) is strictly decreasing: each year multiplies by \(0.9 < 1\), so the value keeps shrinking.
- \(L\) is strictly decreasing: slope \(-4\), the stock falls by 4 units every week.
- \(q\) is neither: it falls until \(x = 6\) (the vertex of the upward parabola) and rises afterwards: there is an optimal machine speed.
Chord rule: connect two graph points and compare with the graph:
- \(T\): concave, the root flattens as \(x\) grows, so chords lie below the graph: diminishing returns per extra worker.
- \(u\): convex, the curve falls steeply first and then flattens toward 8, bending upward; chords lie above the graph.
- \(V\): convex, exponential decay also falls steeply first, then ever flatter; chords lie above the graph.
- \(L\): both convex and concave: for a straight line every chord lies on the graph, so neither inequality is violated.
- \(q\): convex: an upward parabola bends upward everywhere.
Strictly monotone \(\Rightarrow\) one-to-one \(\Rightarrow\) invertible. That covers \(T\), \(u\), \(V\) and \(L\): each output is hit exactly once, so questions like “how many workers for a throughput of 200?” have a unique answer. \(q\) is not one-to-one: \(q(4) = q(8) = 6\), two speeds share one cost: a horizontal line at height \(6\) hits the graph twice, so no inverse exists on all of \(\mathbb{R}\) (only after restricting to \([6, \infty)\) or \((-\infty, 6]\)).
Bounds from the formulas:
- \(u(x) = \frac{600}{x} + 8 > 8\) for every \(x > 0\): the fraction is positive, so 8 is a lower bound: the unit cost never beats €8, however large the batch.
- \(V(t) = 20{,}000 \cdot 0.9^t > 0\): an exponential never reaches zero, so 0 is a lower bound.
- \(q(x) = (x - 6)^2 + 2 \geq 2\): the square is \(\geq 0\), so 2 is a lower bound, attained at \(x = 6\).
- \(T\) is not bounded above: \(40\sqrt{x}\) exceeds any target \(M\) once \(x > (M/40)^2\): concave growth is slow, but it never stops.