Practice Tasks - Session 05-07

Function Determination & Funktionsscharen

Author

Dr. Nikolai Heinrichs & Dr. Tobias Vlćek

Part 1: Quadratic Functions

Problem 1: Three Points (xx)

Find the quadratic function \(f(x) = ax^2 + bx + c\) passing through the points \((0, 3)\), \((1, 6)\), and \((2, 11)\).

Setup: 3 unknowns (\(a, b, c\)), 3 point conditions

Equations:

From \(f(0) = 3\): \(c = 3\)

From \(f(1) = 6\): \(a + b + c = 6\), so \(a + b = 3\)

From \(f(2) = 11\): \(4a + 2b + c = 11\), so \(4a + 2b = 8\), thus \(2a + b = 4\)

Solving:

From \(2a + b = 4\) and \(a + b = 3\):

Subtract: \(a = 1\)

Then: \(b = 3 - 1 = 2\)

Answer: \(f(x) = x^2 + 2x + 3\)

Verification:

  • \(f(0) = 3\) ✓
  • \(f(1) = 1 + 2 + 3 = 6\) ✓
  • \(f(2) = 4 + 4 + 3 = 11\) ✓

Problem 2: Vertex Form (x)

Find the equation of a parabola with vertex at \((-3, 5)\) that passes through the point \((0, -4)\).

Approach: Use vertex form \(f(x) = a(x - h)^2 + k\) where \((h, k)\) is the vertex.

With vertex \((-3, 5)\): \[f(x) = a(x - (-3))^2 + 5 = a(x + 3)^2 + 5\]

Only 1 unknown now: \(a\)

Condition: Passes through \((0, -4)\):

\[f(0) = a(0 + 3)^2 + 5 = -4\] \[9a + 5 = -4\] \[a = -1\]

Answer: \(f(x) = -(x + 3)^2 + 5\)

Expanded form: \(f(x) = -x^2 - 6x - 4\)

Verification: \(f(0) = -9 + 5 = -4\) ✓

Problem 3: Given Maximum (xx)

Find the quadratic function that has a maximum at \((2, 8)\) and passes through the origin.

Setup: \(f(x) = ax^2 + bx + c\) (3 unknowns)

Conditions:

  1. Passes through origin: \(f(0) = 0\)
  2. Maximum at \((2, 8)\): \(f(2) = 8\) (point condition)
  3. Horizontal tangent at max: \(f'(2) = 0\) (derivative condition)

Note: Maximum gives us 2 conditions!

Equations:

From condition 1: \(c = 0\)

\(f'(x) = 2ax + b\)

From condition 3: \(f'(2) = 4a + b = 0\), so \(b = -4a\)

From condition 2: \(4a + 2b + c = 8\)

Substitute \(b = -4a\) and \(c = 0\): \[4a + 2(-4a) + 0 = 8\] \[4a - 8a = 8\] \[-4a = 8\] \[a = -2\]

Then: \(b = -4(-2) = 8\) and \(c = 0\)

Answer: \(f(x) = -2x^2 + 8x\)

Verification:

  • \(f(0) = 0\) ✓
  • \(f(2) = -8 + 16 = 8\) ✓
  • \(f'(2) = -8 + 8 = 0\) ✓
  • \(f''(x) = -4 < 0\) confirms maximum ✓

Problem 4: Point and Slope Conditions (xx)

Find the quadratic function \(f(x) = ax^2 + bx + c\) such that:

  • \(f(1) = 4\)
  • \(f'(1) = 3\)
  • \(f(3) = 6\)

Setup: 3 unknowns, 3 conditions

\(f'(x) = 2ax + b\)

Equations:

  1. \(f(1) = a + b + c = 4\)
  2. \(f'(1) = 2a + b = 3\)
  3. \(f(3) = 9a + 3b + c = 6\)

Solving:

From equation 2: \(b = 3 - 2a\)

Substitute into equation 1: \[a + (3 - 2a) + c = 4\] \[-a + 3 + c = 4\] \[c = 1 + a\]

Substitute both into equation 3: \[9a + 3(3 - 2a) + (1 + a) = 6\] \[9a + 9 - 6a + 1 + a = 6\] \[4a + 10 = 6\] \[a = -1\]

Then:

  • \(b = 3 - 2(-1) = 5\)
  • \(c = 1 + (-1) = 0\)

Answer: \(f(x) = -x^2 + 5x\)

Verification:

  • \(f(1) = -1 + 5 = 4\) ✓
  • \(f'(1) = -2 + 5 = 3\) ✓
  • \(f(3) = -9 + 15 = 6\) ✓

Part 2: Cubic Functions

Problem 5: Four Points (xxx)

Find the cubic function \(f(x) = ax^3 + bx^2 + cx + d\) passing through: \((0, 2)\), \((1, 1)\), \((-1, 3)\), and \((2, 6)\).

Setup: 4 unknowns, 4 point conditions

Equations:

  1. \(f(0) = 2\): \(d = 2\)

  2. \(f(1) = 1\): \(a + b + c + d = 1\), so \(a + b + c = -1\)

  3. \(f(-1) = 3\): \(-a + b - c + d = 3\), so \(-a + b - c = 1\)

  4. \(f(2) = 6\): \(8a + 4b + 2c + d = 6\), so \(8a + 4b + 2c = 4\), thus \(4a + 2b + c = 2\)

System (after substituting \(d = 2\)): \[\begin{cases} a + b + c = -1 \quad \text{...(i)} \\ -a + b - c = 1 \quad \text{...(ii)} \\ 4a + 2b + c = 2 \quad \text{...(iii)} \end{cases}\]

Solving:

Add (i) and (ii): \[2b = 0\] \[b = 0\]

Substitute into (i): \(a + c = -1\), so \(c = -1 - a\) …(iv)

Substitute into (iii): \[4a + 0 + (-1 - a) = 2\] \[3a - 1 = 2\] \[a = 1\]

From (iv): \(c = -1 - 1 = -2\)

Answer: \(f(x) = x^3 - 2x + 2\)

Verification: Check all four points ✓

Problem 6: Two Extrema (xxxx)

Find the cubic function \(f(x) = ax^3 + bx^2 + cx + d\) that has:

  • A local maximum at \((0, 5)\)
  • A local minimum at \((2, 1)\)

Setup: 4 unknowns, 4 conditions (each extremum gives 2)

\(f'(x) = 3ax^2 + 2bx + c\)

Conditions:

  1. \(f(0) = 5\): \(d = 5\)

  2. \(f'(0) = 0\): \(c = 0\)

  3. \(f(2) = 1\): \(8a + 4b + 2c + d = 1\)

  4. \(f'(2) = 0\): \(12a + 4b + c = 0\)

With \(c = 0\) and \(d = 5\):

From condition 3: \(8a + 4b + 5 = 1\), so \(8a + 4b = -4\), thus \(2a + b = -1\) …(i)

From condition 4: \(12a + 4b = 0\), so \(3a + b = 0\) …(ii)

Solving:

From (ii): \(b = -3a\)

Substitute into (i): \[2a + (-3a) = -1\] \[-a = -1\] \[a = 1\]

Then: \(b = -3(1) = -3\)

Answer: \(f(x) = x^3 - 3x^2 + 5\)

Verification:

  • \(f(0) = 5\) ✓
  • \(f'(0) = 0\) ✓
  • \(f(2) = 8 - 12 + 5 = 1\) ✓
  • \(f'(2) = 12 - 12 = 0\) ✓
  • \(f''(x) = 6x - 6\)
    • \(f''(0) = -6 < 0\) → maximum ✓
    • \(f''(2) = 6 > 0\) → minimum ✓

Problem 7: Inflection Point Condition (xxx)

Find the cubic function \(f(x) = ax^3 + bx^2 + cx + d\) such that:

  • \(f(0) = 1\)
  • \(f(1) = 2\)
  • \(f'(0) = 3\)
  • Has an inflection point at \(x = 2\)

Setup: 4 unknowns, 4 conditions

\(f'(x) = 3ax^2 + 2bx + c\)

\(f''(x) = 6ax + 2b\)

Equations:

  1. \(f(0) = 1\): \(d = 1\)

  2. \(f(1) = 2\): \(a + b + c + d = 2\), so \(a + b + c = 1\)

  3. \(f'(0) = 3\): \(c = 3\)

  4. \(f''(2) = 0\): \(12a + 2b = 0\), so \(6a + b = 0\)

From the conditions:

From (1) and (3): \(d = 1\) and \(c = 3\)

From (2): \(a + b + 3 = 1\), so \(a + b = -2\) …(i)

From (4): \(6a + b = 0\) …(ii)

Solving:

From (ii) - (i): \[5a = 2\] \[a = \frac{2}{5}\]

From (i): \(b = -2 - a = -2 - \frac{2}{5} = -\frac{12}{5}\)

Answer: \(f(x) = \frac{2}{5}x^3 - \frac{12}{5}x^2 + 3x + 1\)

Or: \(f(x) = \frac{1}{5}(2x^3 - 12x^2 + 15x + 5)\)

Verification:

  • \(f(0) = 1\) ✓
  • \(f(1) = \frac{2}{5} - \frac{12}{5} + 3 + 1 = \frac{2 - 12 + 15 + 5}{5} = \frac{10}{5} = 2\) ✓
  • \(f'(0) = 3\) ✓
  • \(f''(2) = 12 \cdot \frac{2}{5} - \frac{24}{5} = \frac{24 - 24}{5} = 0\) ✓

Problem 8: Mixed Conditions (xxxx)

Find the cubic function with:

  • Passes through \((1, 4)\)
  • Has a local extremum at \(x = 0\) with \(f(0) = 2\)
  • Has \(f'(2) = 6\)

Setup: \(f(x) = ax^3 + bx^2 + cx + d\) (4 unknowns)

Conditions: Need 4 equations

  1. \(f(1) = 4\) (point)
  2. \(f(0) = 2\) (point at extremum)
  3. \(f'(0) = 0\) (extremum condition)
  4. \(f'(2) = 6\) (slope condition)

\(f'(x) = 3ax^2 + 2bx + c\)

Equations:

  1. \(a + b + c + d = 4\)
  2. \(d = 2\)
  3. \(c = 0\)
  4. \(12a + 4b + c = 6\)

From conditions 2 and 3:

\(d = 2\) and \(c = 0\)

Substitute into remaining:

From equation 1: \(a + b + 0 + 2 = 4\), so \(a + b = 2\) …(i)

From equation 4: \(12a + 4b + 0 = 6\), so \(3a + b = \frac{3}{2}\) …(ii)

Solving:

From (ii) - (i): \[2a = \frac{3}{2} - 2 = -\frac{1}{2}\] \[a = -\frac{1}{4}\]

From (i): \(b = 2 - a = 2 + \frac{1}{4} = \frac{9}{4}\)

Answer: \(f(x) = -\frac{1}{4}x^3 + \frac{9}{4}x^2 + 2\)

Or: \(f(x) = \frac{1}{4}(-x^3 + 9x^2 + 8)\)

Verification: Check all four conditions ✓

Part 3: Funktionsscharen (Function Families with Parameters)

Problem 9: Single Parameter (xx)

For the function family \(f_a(x) = ax^2 - 4ax + 5\), find the value of \(a\) such that \(f_a(3) = 2\).

Condition: \(f_a(3) = 2\)

Substitute: \[a(3)^2 - 4a(3) + 5 = 2\] \[9a - 12a + 5 = 2\] \[-3a = -3\] \[a = 1\]

Answer: \(a = 1\)

Verification: \(f_1(3) = 9 - 12 + 5 = 2\) ✓

Problem 10: Parameter with Extremum (xxx)

For \(g_t(x) = x^3 - 3tx^2 + 4\), find the value(s) of \(t\) such that \(g_t\) has a local extremum at \(x = 2\).

For a local extremum at \(x = 2\): \(g_t'(2) = 0\)

Find the derivative: \[g_t'(x) = 3x^2 - 6tx\]

Set \(g_t'(2) = 0\): \[3(4) - 6t(2) = 0\] \[12 - 12t = 0\] \[t = 1\]

Answer: \(t = 1\)

Classification:

\(g_t''(x) = 6x - 6t\)

For \(t = 1\): \(g_1''(2) = 12 - 6 = 6 > 0\) → local minimum

Verification: \(g_1'(2) = 12 - 12 = 0\) ✓

Problem 11: Two Parameters (xxxx)

Find values of \(a\) and \(b\) such that \(f(x) = x^3 + ax^2 + bx\) has:

  • A local maximum at \(x = 1\) with \(f(1) = 6\)

Setup: 2 unknowns (\(a, b\)), need 2 equations

Conditions:

  1. \(f(1) = 6\) (point condition)
  2. \(f'(1) = 0\) (extremum condition)

\(f'(x) = 3x^2 + 2ax + b\)

Equations:

From condition 1: \[1 + a + b = 6\] \[a + b = 5 \quad \text{...(i)}\]

From condition 2: \[3 + 2a + b = 0\] \[2a + b = -3 \quad \text{...(ii)}\]

Solving:

From (ii) - (i): \[a = -8\]

From (i): \(b = 5 - a = 5 - (-8) = 13\)

Answer: \(a = -8\) and \(b = 13\)

So: \(f(x) = x^3 - 8x^2 + 13x\)

Verification:

  • \(f(1) = 1 - 8 + 13 = 6\) ✓
  • \(f'(1) = 3 - 16 + 13 = 0\) ✓
  • \(f''(x) = 6x - 16\), so \(f''(1) = -10 < 0\) → maximum ✓

Problem 12: Parameter with Two Conditions (xxxx)

For the function family \(h_k(x) = kx^3 - 3kx + 2\), find \(k\) such that:

  • \(h_k\) has a local extremum at \(x = 1\)
  • \(h_k(2) = 0\)

Setup: 1 unknown (\(k\)), 2 conditions

Let’s check if both can be satisfied.

Condition 1: \(h_k'(1) = 0\)

\(h_k'(x) = 3kx^2 - 3k = 3k(x^2 - 1)\)

\(h_k'(1) = 3k(1 - 1) = 0\) for all \(k\)

So \(x = 1\) is always a critical point (regardless of \(k\)).

Condition 2: \(h_k(2) = 0\)

\[k(8) - 3k(2) + 2 = 0\] \[8k - 6k + 2 = 0\] \[2k = -2\] \[k = -1\]

Answer: \(k = -1\)

Verification:

  • \(h_{-1}(x) = -x^3 + 3x + 2\)
  • \(h_{-1}'(1) = 3(-1)(1-1) = 0\) ✓
  • \(h_{-1}(2) = -8 + 6 + 2 = 0\) ✓
  • \(h_{-1}''(x) = -6x + 6\), so \(h_{-1}''(1) = 0\) (inconclusive)
  • Using first derivative test: \(h_{-1}'(x) = -3(x^2-1) = -3(x-1)(x+1)\)
    • Changes from + to - at \(x = 1\) → local maximum ✓

Part 4: Business Applications

Problem 13: Cost from Marginal Cost (xx)

A company’s marginal cost function is \(MC(x) = C'(x) = 6x^2 - 8x + 15\), where \(x\) is thousands of units.

The fixed cost (cost when \(x = 0\)) is €2000.

Find the total cost function \(C(x)\).

Given: \(C'(x) = 6x^2 - 8x + 15\)

Find: \(C(x)\) such that \(C'(x) = 6x^2 - 8x + 15\) and \(C(0) = 2\)

Integration (reverse of differentiation):

Since the derivative of \(x^3\) is \(3x^2\), we have: \[C(x) = 2x^3 - 4x^2 + 15x + k\]

where \(k\) is a constant.

Apply initial condition:

\(C(0) = k = 2\)

Answer: \(C(x) = 2x^3 - 4x^2 + 15x + 2\) (in thousands of euros)

Verification:

  • \(C'(x) = 6x^2 - 8x + 15\) ✓
  • \(C(0) = 2\) ✓

Problem 14: Revenue Function (xxx)

A company knows the following about its revenue function \(R(x) = ax^3 + bx^2 + cx\) (in thousands):

  • Revenue from selling 1000 units (x=1) is €50,000: \(R(1) = 50\)
  • Revenue from selling 2000 units (x=2) is €140,000: \(R(2) = 140\)
  • Marginal revenue at \(x = 1\) is €70,000 per thousand units: \(R'(1) = 70\)

Find the revenue function.

Setup: 3 unknowns (\(a, b, c\)), 3 conditions

\(R'(x) = 3ax^2 + 2bx + c\)

Equations:

  1. \(R(1) = 50\): \(a + b + c = 50\)

  2. \(R(2) = 140\): \(8a + 4b + 2c = 140\), so \(4a + 2b + c = 70\)

  3. \(R'(1) = 70\): \(3a + 2b + c = 70\)

Solving:

From equations 2 and 3: \[(4a + 2b + c) - (3a + 2b + c) = 70 - 70\] \[a = 0\]

From equation 3 with \(a = 0\): \(2b + c = 70\) …(i)

From equation 1 with \(a = 0\): \(b + c = 50\) …(ii)

From (i) - (ii): \(b = 20\)

From (ii): \(c = 50 - 20 = 30\)

Answer: \(R(x) = 20x^2 + 30x\) (in thousands of euros)

Or: \(R(x) = 20000x^2 + 30000x\) euros

Verification:

  • \(R(1) = 20 + 30 = 50\) ✓
  • \(R(2) = 80 + 60 = 140\) ✓
  • \(R'(1) = 40 + 30 = 70\) ✓

Interpretation: Revenue is quadratic (not cubic), suggesting diminishing returns.

Problem 15: Profit Optimization Design (xxxx)

An analyst wants to model a company’s profit function as cubic: \(P(x) = ax^3 + bx^2 + cx + d\) where \(x\) is production level in thousands.

Requirements:

  • Fixed costs (losses when nothing is produced): \(P(0) = -20\) (€20,000 loss)
  • Break-even at 2000 units: \(P(2) = 0\)
  • Maximum profit at 4000 units: Critical point at \(x = 4\)
  • Profit at maximum is €60,000: \(P(4) = 60\)

Find the profit function.

Setup: 4 unknowns, 4 conditions

\(P'(x) = 3ax^2 + 2bx + c\)

Equations:

  1. \(P(0) = -20\): \(d = -20\)

  2. \(P(2) = 0\): \(8a + 4b + 2c + d = 0\)

  3. \(P'(4) = 0\): \(48a + 8b + c = 0\)

  4. \(P(4) = 60\): \(64a + 16b + 4c + d = 60\)

Substitute \(d = -20\):

From equation 2: \(8a + 4b + 2c = 20\), so \(4a + 2b + c = 10\) …(i)

From equation 3: \(48a + 8b + c = 0\) …(ii)

From equation 4: \(64a + 16b + 4c = 80\), so \(16a + 4b + c = 20\) …(iii)

Solving:

From (ii) - (i): \[44a + 6b = -10\] \[22a + 3b = -5 \quad \text{...(iv)}\]

From (iii) - (i): \[12a + 2b = 10\] \[6a + b = 5 \quad \text{...(v)}\]

From (v): \(b = 5 - 6a\)

Substitute into (iv): \[22a + 3(5 - 6a) = -5\] \[22a + 15 - 18a = -5\] \[4a = -20\] \[a = -5\]

Then:

  • \(b = 5 - 6(-5) = 5 + 30 = 35\)
  • From (i): \(4(-5) + 2(35) + c = 10\), so \(-20 + 70 + c = 10\), thus \(c = -40\)

Answer: \(P(x) = -5x^3 + 35x^2 - 40x - 20\) (in thousands of euros)

Verification:

  • \(P(0) = -20\) ✓
  • \(P(2) = -5(8) + 35(4) - 40(2) - 20 = -40 + 140 - 80 - 20 = 0\) ✓
  • \(P'(4) = -15(16) + 70(4) - 40 = -240 + 280 - 40 = 0\) ✓
  • \(P(4) = -5(64) + 35(16) - 40(4) - 20 = -320 + 560 - 160 - 20 = 60\) ✓
  • \(P''(x) = -30x + 70\), so \(P''(4) = -120 + 70 = -50 < 0\) → maximum ✓

Part 5: Advanced Funktionsscharen (Exam Practice)

These problems are typical exam questions. Master them!

Problem 17: Zeros with Discriminant (xxx)

For the function family \(f_t(x) = x^2 + 2tx + t + 6\):

  1. For which values of \(t\) does \(f_t\) have exactly two distinct zeros?

  2. For which values of \(t\) does \(f_t\) have exactly one zero?

  3. For which values of \(t\) does \(f_t\) have no zeros?

Use the discriminant \(\Delta = b^2 - 4ac = (2t)^2 - 4(1)(t+6) = 4t^2 - 4t - 24\)

Factor: \(\Delta = 4(t^2 - t - 6) = 4(t-3)(t+2)\)

  1. Two zeros: \(\Delta > 0\)
    • \((t-3)(t+2) > 0\)
    • Answer: \(t < -2\) or \(t > 3\)
  2. One zero: \(\Delta = 0\)
    • \((t-3)(t+2) = 0\)
    • Answer: \(t = -2\) or \(t = 3\)
  3. No zeros: \(\Delta < 0\)
    • \((t-3)(t+2) < 0\)
    • Answer: \(-2 < t < 3\)

Problem 18: Extremum Location (xxx)

For \(g_t(x) = x^3 - 3tx^2 + 12x\):

  1. Find the value of \(t\) such that \(g_t\) has a local extremum at \(x = 2\).

  2. For that value of \(t\), classify the extremum.

  3. Find \(g_t(2)\) for that value of \(t\).

\(g_t'(x) = 3x^2 - 6tx + 12\)

  1. Need \(g_t'(2) = 0\):
    • \(3(4) - 6t(2) + 12 = 0\)
    • \(12 - 12t + 12 = 0\)
    • \(24 = 12t\)
    • Answer: \(t = 2\)
  2. \(g_2''(x) = 6x - 12\)
    • \(g_2''(2) = 12 - 12 = 0\) (inconclusive!)
    • Use first derivative test: \(g_2'(x) = 3x^2 - 12x + 12 = 3(x-2)^2\)
    • Since \((x-2)^2 \geq 0\) always, \(g_2'(x) \geq 0\) always
    • No sign change → Neither max nor min (inflection point with horizontal tangent)
  3. \(g_2(2) = 8 - 12(2) + 24 = 8 - 24 + 24 = 8\)

Problem 19: Inflection Point Parameter (xxx)

For \(h_t(x) = x^3 + tx^2 - 9x + 5\):

  1. Find \(t\) such that \(h_t\) has an inflection point at \(x = 1\).

  2. For that \(t\), find the coordinates of the inflection point.

\(h_t'(x) = 3x^2 + 2tx - 9\)

\(h_t''(x) = 6x + 2t\)

  1. For inflection at \(x = 1\): \(h_t''(1) = 0\)
    • \(6(1) + 2t = 0\)
    • \(6 + 2t = 0\)
    • Answer: \(t = -3\)
  2. With \(t = -3\): \(h_{-3}(x) = x^3 - 3x^2 - 9x + 5\)
    • \(h_{-3}(1) = 1 - 3 - 9 + 5 = -6\)
    • Inflection point: \((1, -6)\)

Verification: Check concavity changes:

  • \(h_{-3}''(x) = 6x - 6\)
  • \(h_{-3}''(x) < 0\) for \(x < 1\) (concave down)
  • \(h_{-3}''(x) > 0\) for \(x > 1\) (concave up)
  • Concavity changes ✓

Problem 20: Function Value Condition (xx)

For \(f_k(x) = kx^2 - 4x + k\):

  1. Find \(k\) such that \(f_k(1) = 3\).

  2. For that \(k\), find the vertex of the parabola.

  3. Does \(f_k\) open upward or downward?

  1. \(f_k(1) = k - 4 + k = 2k - 4 = 3\)
    • \(2k = 7\)
    • Answer: \(k = 3.5\)
  2. With \(k = 3.5\): \(f_{3.5}(x) = 3.5x^2 - 4x + 3.5\)
    • Vertex x-coordinate: \(x = \frac{4}{2(3.5)} = \frac{4}{7}\)
    • Vertex y-coordinate: \(f_{3.5}(\frac{4}{7}) = 3.5(\frac{16}{49}) - 4(\frac{4}{7}) + 3.5\)
    • \(= \frac{56}{49} - \frac{16}{7} + 3.5 = \frac{8}{7} - \frac{16}{7} + \frac{24.5}{7} = \frac{16.5}{7} \approx 2.36\)
    • Vertex: \((\frac{4}{7}, \frac{33}{14})\) or approximately \((0.57, 2.36)\)
  3. Since \(k = 3.5 > 0\), the parabola opens upward

Problem 21: Two Zeros at Specific Points (xxxx)

For \(g_t(x) = x^2 - tx + t - 3\):

Find \(t\) such that \(g_t\) has zeros at \(x = 1\) and \(x = 3\).

Method 1: Direct substitution

If \(x = 1\) is a zero: \(g_t(1) = 1 - t + t - 3 = -2 \neq 0\)

This is impossible! Let’s verify with \(x = 3\): \(g_t(3) = 9 - 3t + t - 3 = 6 - 2t\)

For zero: \(6 - 2t = 0 \implies t = 3\)

Check \(x = 1\) with \(t = 3\): \(g_3(1) = 1 - 3 + 3 - 3 = -2 \neq 0\)

Answer: No value of \(t\) exists! The condition is impossible.

Why? By Vieta’s formulas, if zeros are 1 and 3:

  • Sum of zeros = \(1 + 3 = 4 = t\) (coefficient of \(x\))
  • Product of zeros = \(1 \times 3 = 3 = t - 3\) (constant term)

From product: \(t = 6\), but from sum: \(t = 4\). Contradiction!

This is a classic exam trap - always check if conditions are consistent.

Problem 22: Maximum Value Parameter (xxx)

For \(p_t(x) = -x^2 + 4x + t\):

  1. What is the maximum value of \(p_t\) (in terms of \(t\))?

  2. Find \(t\) such that the maximum value is 10.

  3. For that \(t\), find the zeros of \(p_t\).

  1. This is a downward parabola. Maximum at vertex.
    • Vertex x-coordinate: \(x = \frac{-4}{2(-1)} = 2\)
    • Maximum value: \(p_t(2) = -4 + 8 + t = 4 + t\)
    • Maximum value: \(4 + t\)
  2. Set \(4 + t = 10\)
    • Answer: \(t = 6\)
  3. With \(t = 6\): \(p_6(x) = -x^2 + 4x + 6\)
    • Using quadratic formula: \(x = \frac{-4 \pm \sqrt{16 + 24}}{-2} = \frac{-4 \pm \sqrt{40}}{-2}\)
    • \(= \frac{-4 \pm 2\sqrt{10}}{-2} = 2 \mp \sqrt{10}\)
    • Zeros: \(x = 2 - \sqrt{10} \approx -1.16\) and \(x = 2 + \sqrt{10} \approx 5.16\)