Business Calculus Exam 2 Review – Step-by-Step Guidance

Q1(a). Find the derivative of $40e^{-2x}$ with respect to $x$.

Background

Topic: Differentiation of Exponential Functions

This question tests your ability to differentiate exponential functions, specifically those involving the chain rule.

Key Terms and Formulas:

• Exponential function: $e^{u(x)}$

• Derivative: $\frac{d}{dx}e^{u(x)} = e^{u(x)} \cdot u'(x)$

• Chain Rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

Step-by-Step Guidance

1. Identify the outer function ($e^{u(x)}$) and the inner function ($u(x) = -2x$).

2. Differentiate the outer function with respect to the inner function: $\frac{d}{dx}e^{u(x)} = e^{u(x)} \cdot u'(x)$.

3. Find the derivative of the inner function $u(x) = -2x$.

4. Multiply the constant 40 by the derivative you found in the previous steps.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}[40e^{-2x}] = 40 \cdot e^{-2x} \cdot (-2) = -80e^{-2x}$

The chain rule gives us the extra factor of $-2$ from differentiating $-2x$.

Q1(b). Find the derivative of $\ln|t^3 + 4t^2 + 2|$ with respect to $t$.

Background

Topic: Differentiation of Logarithmic Functions

This question tests your ability to differentiate natural logarithms of composite functions using the chain rule.

Key Terms and Formulas:

• Derivative: $\frac{d}{dt}\ln|u(t)| = \frac{u'(t)}{u(t)}$

• Chain Rule: $\frac{d}{dt}f(g(t)) = f'(g(t)) \cdot g'(t)$

Step-by-Step Guidance

1. Let $u(t) = t^3 + 4t^2 + 2$.

2. Find $u'(t)$, the derivative of $u(t)$ with respect to $t$.

3. Apply the formula for the derivative of $\ln|u(t)|$.

4. Write the result as a single fraction.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dt}[\ln|t^3 + 4t^2 + 2|] = \frac{3t^2 + 8t}{t^3 + 4t^2 + 2}$

We used the chain rule and the derivative of the natural logarithm function.

Q1(c). Find the derivative of $\sqrt{x^4 + 5}$ with respect to $x$.

Background

Topic: Differentiation using the Chain Rule

This question tests your ability to differentiate composite functions, specifically a square root of a polynomial.

Key Terms and Formulas:

• Rewrite $\sqrt{x^4 + 5}$ as $(x^4 + 5)^{1/2}$

• Power Rule: $\frac{d}{dx}u^n = n u^{n-1} u'(x)$

• Chain Rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

Step-by-Step Guidance

1. Let $u(x) = x^4 + 5$ and rewrite the function as $u(x)^{1/2}$.

2. Apply the power rule: bring down the exponent and subtract one from it.

3. Find $u'(x)$, the derivative of $x^4 + 5$.

4. Multiply the results from the previous steps together.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}[\sqrt{x^4 + 5}] = \frac{1}{2}(x^4 + 5)^{-1/2} \cdot 4x^3 = \frac{2x^3}{\sqrt{x^4 + 5}}$

We used the chain rule and the power rule for differentiation.

Q1(d). Find the derivative of $(5x + 3)^4 e^{x^3}$ with respect to $x$.

Background

Topic: Product Rule, Chain Rule, and Exponential Functions

This question tests your ability to use the product rule, chain rule, and differentiate exponential and polynomial functions.

Key Terms and Formulas:

• Product Rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$

• Chain Rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

• Exponential Derivative: $\frac{d}{dx}e^{u(x)} = e^{u(x)} u'(x)$

Step-by-Step Guidance

1. Let $u(x) = (5x + 3)^4$ and $v(x) = e^{x^3}$.

2. Find $u'(x)$ using the chain rule and power rule.

3. Find $v'(x)$ using the chain rule for the exponential function.

4. Apply the product rule: $u'(x)v(x) + u(x)v'(x)$.

5. Write out the full expression, but do not simplify completely yet.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{d}{dx}[(5x + 3)^4 e^{x^3}] = 4(5x + 3)^3 \cdot 5 \cdot e^{x^3} + (5x + 3)^4 \cdot e^{x^3} \cdot 3x^2$

Or, factoring $e^{x^3}$: $e^{x^3} [20(5x + 3)^3 + 3x^2(5x + 3)^4]$

Q2(a). For $f(x) = \frac{2}{3}x^3 + x^2 - 12x + 10$, find the intervals where the function is increasing and decreasing, and identify local maxima and minima.

Background

Topic: First Derivative Test

This question tests your ability to use the first derivative to determine where a function increases or decreases and to find local extrema.

Key Terms and Formulas:

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

• First Derivative Test: Analyze sign changes of $f'(x)$ around critical points

Step-by-Step Guidance

1. Find $f'(x)$, the first derivative of $f(x)$.

2. Set $f'(x) = 0$ and solve for $x$ to find critical points.

3. Make a sign chart for $f'(x)$ around the critical points to determine intervals of increase/decrease.

4. Use the sign changes to identify local maxima and minima.

Try solving on your own before revealing the answer!

Final Answer:

$f'(x) = 2x^2 + 2x - 12$

Critical points at $x = -3$ and $x = 2$.

Increasing on $(-\infty, -3)$ and $(2, \infty)$; decreasing on $(-3, 2)$.

Local maximum at $x = -3$, local minimum at $x = 2$.

Q2(b). For $g(x) = x^3 e^{-4x}$, find the intervals where the function is increasing and decreasing, and identify local maxima and minima.

Background

Topic: First Derivative Test with Product Rule

This question tests your ability to use the product rule and first derivative test to analyze increasing/decreasing intervals and local extrema.

Key Terms and Formulas:

• Product Rule: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$

• First Derivative Test

Step-by-Step Guidance

1. Let $u(x) = x^3$ and $v(x) = e^{-4x}$; find $u'(x)$ and $v'(x)$.

2. Apply the product rule to find $g'(x)$.

3. Set $g'(x) = 0$ and solve for $x$ to find critical points.

4. Make a sign chart for $g'(x)$ to determine intervals of increase/decrease and identify local extrema.

Try solving on your own before revealing the answer!

Final Answer:

$g'(x) = 3x^2 e^{-4x} - 4x^3 e^{-4x} = x^2 e^{-4x}(3 - 4x)$

Critical points at $x = 0$ and $x = 0.75$.

Increasing on $(0, 0.75)$, decreasing on $(0.75, \infty)$; local maximum at $x = 0.75$.

Q3(a). Let $f(x) = x^3 - 9x^2 + 4x + 6$. Find $f'(x)$, $f''(x)$, $f'''(x)$, and $f^{(4)}(x)$.

Background

Topic: Higher-Order Derivatives

This question tests your ability to compute successive derivatives of a polynomial function.

Key Terms and Formulas:

• Power Rule: $\frac{d}{dx}x^n = n x^{n-1}$

• Higher Derivatives: Take the derivative repeatedly

Step-by-Step Guidance

1. Find $f'(x)$ by differentiating $f(x)$ once.

2. Find $f''(x)$ by differentiating $f'(x)$.

3. Find $f'''(x)$ by differentiating $f''(x)$.

4. Find $f^{(4)}(x)$ by differentiating $f'''(x)$.

Try solving on your own before revealing the answer!

Final Answer:

$f'(x) = 3x^2 - 18x + 4$

$f''(x) = 6x - 18$

$f'''(x) = 6$

$f^{(4)}(x) = 0$

Q3(b). Let $g(x) = \ln|5x + 1|$. Find $g''(x)$.

Background

Topic: Higher-Order Derivatives of Logarithmic Functions

This question tests your ability to find the second derivative of a logarithmic function using the chain rule.

Key Terms and Formulas:

• First derivative: $\frac{d}{dx}\ln|u(x)| = \frac{u'(x)}{u(x)}$

• Second derivative: Differentiate the first derivative using the quotient rule

Step-by-Step Guidance

1. Let $u(x) = 5x + 1$; find $g'(x)$ using the chain rule.

2. Differentiate $g'(x)$ to find $g''(x)$, applying the quotient rule as needed.

3. Simplify the resulting expression for $g''(x)$.

Try solving on your own before revealing the answer!

Final Answer:

$g'(x) = \frac{5}{5x + 1}$

$g''(x) = -\frac{25}{(5x + 1)^2}$

Q4. For $f(x) = 2x^3 - 15x^2 + 24x + 6$, find the intervals where the function is concave up or down, and identify inflection points.

Background

Topic: Concavity and Inflection Points

This question tests your ability to use the second derivative to determine concavity and locate inflection points.

Key Terms and Formulas:

• Second Derivative: $f''(x)$

• Concave up: $f''(x) > 0$; Concave down: $f''(x) < 0$

• Inflection point: Where $f''(x)$ changes sign

Step-by-Step Guidance

1. Find $f''(x)$ by differentiating $f(x)$ twice.

2. Set $f''(x) = 0$ and solve for $x$ to find possible inflection points.

3. Make a sign chart for $f''(x)$ to determine intervals of concavity.

4. Identify the $x$-values where concavity changes (inflection points).

Try solving on your own before revealing the answer!

Final Answer:

$f''(x) = 12x - 30$

Inflection point at $x = 2.5$

Concave up on $(2.5, \infty)$, concave down on $(-\infty, 2.5)$

Q5. For $f(x) = \frac{1}{4}x^4 - \frac{7}{3}x^3 + 5x^2 - 2$ on $[-1, 4]$, find the absolute maximum and minimum values.

Background

Topic: Absolute Extrema on a Closed Interval

This question tests your ability to find absolute maximum and minimum values of a function on a closed interval using critical points and endpoints.

Key Terms and Formulas:

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

• Evaluate $f(x)$ at critical points and endpoints

Step-by-Step Guidance

1. Find $f'(x)$ and set it equal to zero to find critical points inside $[-1, 4]$.

2. Evaluate $f(x)$ at each critical point and at the endpoints $x = -1$ and $x = 4$.

3. Compare all values to determine the absolute maximum and minimum.

Try solving on your own before revealing the answer!

Final Answer:

Critical points at $x = 1$ and $x = 6$ (only $x = 1$ is in $[-1, 4]$).

Evaluate $f(x)$ at $x = -1$, $x = 1$, and $x = 4$:

• $f(-1) = -\frac{1}{4} + \frac{7}{3} + 5 - 2 \approx 5.75$

• $f(1) = \frac{1}{4} - \frac{7}{3} + 5 - 2 \approx 0.25$

• $f(4) = 64/4 - 448/3 + 80 - 2 \approx -38.67$

Absolute maximum at $x = -1$ ($f(-1) \approx 5.75$), absolute minimum at $x = 4$ ($f(4) \approx -38.67$).

Q6(a). The cost of producing $x$ units is $C(x) = 300 + 1.2x$. The price corresponding to a demand for $x$ units is $p = D(x) = 200 - 0.2x$. Find the quantity $x$ and price $p$ that would maximize profits.

Background

Topic: Optimization – Maximizing Profit

This question tests your ability to set up and solve an optimization problem involving cost, revenue, and profit functions.

Key Terms and Formulas:

• Revenue: $R(x) = p \cdot x$

• Profit: $P(x) = R(x) - C(x)$

• To maximize profit, set $P'(x) = 0$ and solve for $x$

Step-by-Step Guidance

1. Write the revenue function $R(x)$ using the demand function $p = D(x)$.

2. Write the profit function $P(x) = R(x) - C(x)$.

3. Find $P'(x)$ and set it equal to zero to find the value of $x$ that maximizes profit.

4. Plug this value of $x$ back into the demand function to find the corresponding price $p$.

Try solving on your own before revealing the answer!

Final Answer:

Profit is maximized at $x = 497.5$ units.

Corresponding price: $p = 200 - 0.2(497.5) = 100.5$ dollars per unit.

Q6(b). A farmer needs to build a rectangular pen with one side against a stream (no fencing needed on that side). Fencing costs $10 per meter$, and the budget is $1500$. What is the maximum area that can be enclosed?

Background

Topic: Optimization – Area with Constraints

This question tests your ability to maximize area given a cost constraint, using calculus and algebraic manipulation.

Key Terms and Formulas:

• Let $x$ = length parallel to the stream, $y$ = width (perpendicular)

• Perimeter (fenced): $x + 2y$ (since one side is against the stream)

• Cost constraint: $10(x + 2y) = 1500$

• Area: $A = x \cdot y$

• Express $A$ in terms of one variable using the constraint, then maximize $A$

Step-by-Step Guidance

1. Write the cost constraint equation and solve for one variable in terms of the other.

2. Express the area $A$ as a function of a single variable using the constraint.

3. Find $A'(x)$ and set it equal to zero to find the value of $x$ that maximizes area.

4. Use this value to find the corresponding $y$ and calculate the maximum area.

Try solving on your own before revealing the answer!

Final Answer:

Maximum area is $5625$square meters, with dimensions$x = 750$meters (parallel to stream),$y = 7.5$ meters (width).

This uses all the fencing budget and maximizes the enclosed area.