Business Calculus Final Exam Review: Step-by-Step Guidance

Q1. Find the equation of the secant line from the graph to the right.

Background

Topic: Secant Lines and Average Rate of Change

This question tests your understanding of how to find the equation of a secant line, which connects two points on a function's graph and represents the average rate of change between those points.

Key Terms and Formulas:

• Secant Line: A line passing through two points on a curve.

• Average Rate of Change: $m = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

• Point-Slope Form: $y - y_1 = m(x - x_1)$

Step-by-Step Guidance

1. Identify the two points on the graph where the secant line will pass. These are usually marked or specified in the question.

2. Find the coordinates of these points: $(x_1, y_1)$ and $(x_2, y_2)$.

3. Calculate the slope $m$ using $m = \frac{y_2 - y_1}{x_2 - x_1}$.

4. Write the equation of the secant line using the point-slope form: $y - y_1 = m(x - x_1)$.

Try solving on your own before revealing the answer!

Final Answer:

The equation of the secant line is $y = m x + b$, where $m$ is the slope calculated from the two points and $b$ is found using one of the points.

For example, if the points are $(a, f(a))$ and $(b, f(b))$, then $m = \frac{f(b) - f(a)}{b - a}$ and $b = f(a) - m a$.

Q2. Given the table below, calculate the following derivatives:

Background

Topic: Product and Quotient Rule for Derivatives

This question tests your ability to use the product and quotient rules to find derivatives using values from a table.

Key Terms and Formulas:

• Product Rule: $[f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)$

• Quotient Rule: $\left[\frac{f(x)}{g(x)}\right]' = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

Step-by-Step Guidance

1. For $p(x) = f(x)g(x)$, use the product rule: $p'(x) = f'(x)g(x) + f(x)g'(x)$.

2. Locate the values of $f(x)$, $f'(x)$, $g(x)$, and $g'(x)$ for the specified $x$ in the table.

3. Plug these values into the product rule formula for the required $x$ (e.g., $x=2$).

4. For $q(x) = \frac{f(x)}{g(x)}$, use the quotient rule: $q'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$.

5. Locate the values for the specified $x$ and substitute them into the quotient rule formula.

Try solving on your own before revealing the answer!

Final Answer:

For $p'(2)$, substitute the values from the table into $p'(x) = f'(x)g(x) + f(x)g'(x)$.

For $q'(1)$, substitute the values from the table into $q'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$.

Q3. Find the critical values for $f(x) = 2x^3 - 3x^2 - 36x$ and use the First and Second Derivative Test to assess if a critical value is an extrema for $f(x) = x e^{-3x}$.

Background

Topic: Critical Points and Extrema

This question tests your ability to find critical points by setting the first derivative to zero and then use the first and second derivative tests to determine if those points are maxima, minima, or neither.

Key Terms and Formulas:

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

• First Derivative Test: Examines sign changes in $f'(x)$ around critical points.

• Second Derivative Test: Uses $f''(x)$ at critical points to determine concavity.

Step-by-Step Guidance

1. Find $f'(x)$ for $f(x) = 2x^3 - 3x^2 - 36x$.

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

3. For $f(x) = x e^{-3x}$, find $f'(x)$ and $f''(x)$.

4. Use the first derivative test: Check the sign of $f'(x)$ before and after each critical point.

5. Use the second derivative test: Evaluate $f''(x)$ at each critical point to determine if it is a maximum or minimum.

Try solving on your own before revealing the answer!

Final Answer:

Critical values are found by solving $f'(x) = 0$. Use the first and second derivative tests to classify each critical value as a local maximum, minimum, or neither.

For $f(x) = x e^{-3x}$, check the sign of $f'(x)$ and the value of $f''(x)$ at each critical point.