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 & 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 that the secant line will pass through. 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 secant line equation is $y = m x + b$, where $m$ is the slope calculated from the two points and $b$ is found by plugging in 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 $b = f(a) - m a$.

This gives the full equation for the secant line.

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

Background

Topic: Derivatives from Tables (Product and Quotient Rule)

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

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. Look up the values of $f(x)$, $f'(x)$, $g(x)$, and $g'(x)$ in the table for the specified $x$ value.

3. Plug these values into the formula and simplify, but stop before the final calculation.

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. Again, look up the necessary values in the table and set up the expression for $q'(x)$, but do not compute the final value.

Try solving on your own before revealing the answer!

Final Answer:

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

For $q'(1)$, substitute: $q'(1) = \frac{f'(1)g(1) - f(1)g'(1)}{[g(1)]^2}$.

For $h'(4)$, if $h(x) = g(f(x))$, use the chain rule: $h'(x) = g'(f(x)) \cdot f'(x)$, and substitute the values from the table.

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 $f(x) = x e^{-3x}$. Justify.

Background

Topic: Critical Points, Extrema, and Derivative Tests

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 local maxima, minima, or neither.

Key Terms and Formulas:

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

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

• Second Derivative Test: If $f''(x) > 0$ at a critical point, it's a minimum; if $f''(x) < 0$, it's a maximum.

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 values.

3. For $f(x) = x e^{-3x}$, find $f'(x)$ and set it to zero to find critical points.

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

5. Use the second derivative test: Compute $f''(x)$ at each critical value to determine if it's a max or min.

Try solving on your own before revealing the answer!

Final Answer:

Critical values for $f(x) = 2x^3 - 3x^2 - 36x$ are found by solving $f'(x) = 6x^2 - 6x - 36 = 0$.

For $f(x) = x e^{-3x}$, use the product rule to find $f'(x)$, set it to zero, and apply the first and second derivative tests as described.

Justification comes from checking the sign of $f'(x)$ and the value of $f''(x)$ at each critical point.