Business Calculus Exam 4 Study Guide – Step-by-Step Guidance

Q1. Find the absolute extrema of $f(x) = 5 \ln(x) - 4x$ on $[1,3]$.

Background

Topic: Absolute Extrema (Maxima and Minima) on a Closed Interval

This question tests your ability to find the absolute maximum and minimum values of a function on a closed interval using calculus techniques.

Key Terms and Formulas:

• Critical Number: A value $c$ in the domain of $f$ where $f'(c) = 0$ or $f'(c)$ does not exist.

• Absolute Maximum/Minimum: The largest/smallest value of $f(x)$ on $[a,b]$.

• Procedure: Check endpoints and critical points inside the interval.

• Derivative: $f'(x)$ is needed to find critical points.

Step-by-Step Guidance

1. Check that $f(x)$ is continuous on $[1,3]$. Since $\ln(x)$ is continuous for $x>0$, $f(x)$ is continuous on $[1,3]$.

2. Find the derivative: $f'(x) = \frac{5}{x} - 4$.

3. Set $f'(x) = 0$ to find critical points inside $(1,3)$: $\frac{5}{x} - 4 = 0$.

4. Solve for $x$ to find the critical number in $(1,3)$.

5. Evaluate $f(x)$ at the endpoints $x=1$ and $x=3$, and at the critical number found in the previous step.

Try solving on your own before revealing the answer!

Final Answer:

Critical number $x=\frac{5}{4}$ on $[1,3]$.

Absolute maximum is approximately $-3.884$ at $x=\frac{5}{4}$ and absolute minimum is approximately $-6.507$ at $x=3$.

We evaluated $f(x)$ at the endpoints and the critical point to determine the extrema.

Q2. Find the absolute extrema of $f(x) = x^4 - 6x^2$ on:

• a. $[1,3]$

• b. $[-2,0]$

Background

Topic: Absolute Extrema on a Closed Interval

This question tests your ability to find absolute maximum and minimum values for a polynomial function on given intervals.

Key Terms and Formulas:

• Find $f'(x)$, set to zero to find critical points.

• Evaluate $f(x)$ at endpoints and critical points within the interval.

Step-by-Step Guidance

1. Find the derivative: $f'(x) = 4x^3 - 12x$.

2. Set $f'(x) = 0$ and solve for $x$ to find critical points: $4x^3 - 12x = 0$.

3. Factor and solve: $4x(x^2 - 3) = 0$ gives $x=0$, $x=\sqrt{3}$, $x=-\sqrt{3}$.

4. For each interval, determine which critical points are within the interval.

5. Evaluate $f(x)$ at the endpoints and at the critical points within the interval.

Try solving on your own before revealing the answer!

Final Answer:

a. On $[1,3]$: Absolute maximum is $27$at$x=3$, absolute minimum is$-9$ at $x=\sqrt{3}$.

b. On $[-2,0]$: Absolute maximum is $0$at$x=0$, absolute minimum is$-9$ at $x=-\sqrt{3}$.

We checked all endpoints and critical points within each interval.

Q3. Find the absolute extrema of $f(x) = x^3 - 27x + 1$ on $[-4,0]$.

Background

Topic: Absolute Extrema on a Closed Interval

This question tests your ability to find absolute maximum and minimum values for a cubic function on a closed interval.

Key Terms and Formulas:

• Find $f'(x)$, set to zero to find critical points.

• Evaluate $f(x)$ at endpoints and critical points within the interval.

Step-by-Step Guidance

1. Find the derivative: $f'(x) = 3x^2 - 27$.

2. Set $f'(x) = 0$ and solve for $x$: $3x^2 - 27 = 0$.

3. Solve for $x$ to find critical points: $x^2 = 9 \implies x = 3, -3$.

4. Determine which critical points are in $[-4,0]$ (only $x=-3$).

5. Evaluate $f(x)$ at $x=-4$, $x=-3$, and $x=0$.

Try solving on your own before revealing the answer!

Final Answer:

Critical number $x=-3$ is in $[-4,0]$.

Absolute maximum is $55$at$x=-3$, absolute minimum is $1$ at $x=0$.

We evaluated $f(x)$ at the endpoints and the critical point within the interval.

Q4. Find the absolute extrema of $f(x) = \frac{2x}{x+3}$ on $[1,4]$.

Background

Topic: Absolute Extrema for Rational Functions

This question tests your ability to find absolute maximum and minimum values for a rational function on a closed interval.

Key Terms and Formulas:

• Find $f'(x)$, set to zero to find critical points.

• Evaluate $f(x)$ at endpoints and critical points within the interval.

Step-by-Step Guidance

1. Find the derivative using the quotient rule: $f'(x) = \frac{(x+3)\cdot2 - 2x\cdot1}{(x+3)^2}$.

2. Simplify $f'(x)$ and set it to zero to find critical points.

3. Solve for $x$ to find critical numbers in $(1,4)$.

4. Evaluate $f(x)$ at $x=1$, $x=4$, and at the critical number found.

Try solving on your own before revealing the answer!

Final Answer:

Absolute minimum is $\frac{1}{2}$ at $x=1$ and absolute maximum is $\frac{8}{7}$ at $x=4$.

We used the quotient rule to find the critical point and evaluated $f(x)$ at all relevant points.