Absolute Extrema on Closed Intervals – Business Calculus Guidance

Q1. Find the absolute maximum and absolute minimum of $f(x) = 9 - x^2$ on $[-4, 4]$.

Background

Topic: Absolute Extrema on Closed Intervals

This question tests your ability to find the absolute maximum and minimum values of a function on a closed interval. This is a fundamental concept in calculus, especially for optimization problems in business contexts.

Key Terms and Formulas:

• Absolute Maximum: The largest value of $f(x)$ on the interval.

• Absolute Minimum: The smallest value of $f(x)$ on the interval.

• Critical Points: Points where $f'(x) = 0$ or $f'(x)$ is undefined.

• Endpoints: The values of $x$ at the boundaries of the interval.

To find absolute extrema on a closed interval, evaluate $f(x)$ at critical points and endpoints.

Step-by-Step Guidance

1. Find the derivative: $f'(x) = \frac{d}{dx}(9 - x^2) = -2x$.

2. Set $f'(x) = 0$ to find critical points: $-2x = 0 \Rightarrow x = 0$.

3. List all candidates for extrema: critical points ($x = 0$) and endpoints ($x = -4$, $x = 4$).

4. Set up $f(x)$ for each candidate: $f(-4)$, $f(0)$, $f(4)$.

Try solving on your own before revealing the answer!

Final Answer:

$f(-4) = 9 - (-4)^2 = 9 - 16 = -7$

$f(0) = 9 - 0^2 = 9$

$f(4) = 9 - 4^2 = 9 - 16 = -7$

The absolute maximum is $9$at$x = 0$, and the absolute minimum is$-7$ at $x = -4$and$x = 4$.

Q2. Find the absolute maximum and absolute minimum of $f(x) = x^2 - 6x + 7$ on:

• A) $[0, 10]$

• B) $[4, 10]$

Background

Topic: Absolute Extrema of Quadratic Functions

This question asks you to find the absolute maximum and minimum values of a quadratic function on two different intervals. Quadratic functions are common in business applications, such as cost and revenue models.

Key Terms and Formulas:

• Critical Points: Where $f'(x) = 0$.

• Endpoints: The values at the boundaries of the interval.

• Derivative: $f'(x) = \frac{d}{dx}(x^2 - 6x + 7) = 2x - 6$.

Step-by-Step Guidance

1. Find the derivative: $f'(x) = 2x - 6$.

2. Set $f'(x) = 0$ to find critical points: $2x - 6 = 0 \Rightarrow x = 3$.

3. For each interval, check if the critical point $x = 3$ is within the interval.

4. List candidates for extrema for each interval:

   • A) $x = 0$, $x = 3$, $x = 10$

   • B) $x = 4$, $x = 10$ (since $x = 3$ is not in $[4, 10]$)

5. Set up $f(x)$ for each candidate in both intervals.

Try solving on your own before revealing the answer!

Final Answer:

A) $[0, 10]$:

$f(0) = 0^2 - 6 \cdot 0 + 7 = 7$

$f(3) = 3^2 - 6 \cdot 3 + 7 = 9 - 18 + 7 = -2$

$f(10) = 10^2 - 6 \cdot 10 + 7 = 100 - 60 + 7 = 47$

Absolute minimum: $-2$$at$; absolute maximum: $47$ at $x = 10$.

B) $[4, 10]$:

$f(4) = 4^2 - 6 \cdot 4 + 7 = 16 - 24 + 7 = -1$

$f(10) = 47$ (as above)

Absolute minimum: $-1$$at$; absolute maximum: $47$ at $x = 10$.

Q3. Find the absolute maximum and absolute minimum of $f(x) = x^3 - 6x^2 + 9x - 6$ on $[-1, 5]$.

Background

Topic: Absolute Extrema of Cubic Functions

This question tests your ability to find absolute extrema for a cubic function on a closed interval. Cubic functions can model more complex business scenarios, such as profit functions with inflection points.

Key Terms and Formulas:

• Critical Points: Where $f'(x) = 0$.

• Endpoints: The values at the boundaries of the interval.

• Derivative: $f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x - 6) = 3x^2 - 12x + 9$.

Step-by-Step Guidance

1. Find the derivative: $f'(x) = 3x^2 - 12x + 9$.

2. Set $f'(x) = 0$ to find critical points: $3x^2 - 12x + 9 = 0$.

3. Solve the quadratic equation for $x$ to find critical points.

4. Check which critical points are within $[-1, 5]$.

5. List candidates for extrema: endpoints ($x = -1$, $x = 5$) and valid critical points.

6. Set up $f(x)$ for each candidate.

Try solving on your own before revealing the answer!

Final Answer:

Solving $3x^2 - 12x + 9 = 0$ gives $x = 1$ and $x = 3$.

Evaluate $f(x)$ at $x = -1$, $x = 1$, $x = 3$, $x = 5$:

$f(-1) = (-1)^3 - 6(-1)^2 + 9(-1) - 6 = -1 - 6 + -9 - 6 = -22$

$f(1) = 1 - 6 + 9 - 6 = -2$

$f(3) = 27 - 54 + 27 - 6 = -6$

$f(5) = 125 - 150 + 45 - 6 = 14$

Absolute maximum: $14$at$x = 5$; absolute minimum:$-22$ at $x = -1$.

Q4. Find the absolute maximum and absolute minimum of $f(x) = 2x^3 - 3x^2 - 12x + 1$ on $[-3, 1]$.

Background

Topic: Absolute Extrema of Cubic Functions

This question tests your ability to find absolute extrema for a cubic function on a closed interval, which is important for optimization in business calculus.

Key Terms and Formulas:

• Critical Points: Where $f'(x) = 0$.

• Endpoints: The values at the boundaries of the interval.

• Derivative: $f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 1) = 6x^2 - 6x - 12$.

Step-by-Step Guidance

1. Find the derivative: $f'(x) = 6x^2 - 6x - 12$.

2. Set $f'(x) = 0$ to find critical points: $6x^2 - 6x - 12 = 0$.

3. Solve the quadratic equation for $x$ to find critical points.

4. Check which critical points are within $[-3, 1]$.

5. List candidates for extrema: endpoints ($x = -3$, $x = 1$) and valid critical points.

6. Set up $f(x)$ for each candidate.

Try solving on your own before revealing the answer!

Final Answer:

Solving $6x^2 - 6x - 12 = 0$ gives $x = -1$ and $x = 2$.

Only $x = -1$ is in $[-3, 1]$.

Evaluate $f(x)$ at $x = -3$, $x = -1$, $x = 1$:

$f(-3) = 2(-3)^3 - 3(-3)^2 - 12(-3) + 1 = 2(-27) - 3(9) + 36 + 1 = -54 - 27 + 36 + 1 = -44$

$f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 1 = 2(-1) - 3(1) + 12 + 1 = -2 - 3 + 12 + 1 = 8$

$f(1) = 2(1)^3 - 3(1)^2 - 12(1) + 1 = 2 - 3 - 12 + 1 = -12$

Absolute maximum: $8$at$x = -1$; absolute minimum:$-44$ at $x = -3$.