Calculus $1$ : Exam $2$ $\newline$ Name: $\newline$ $15$ . ( $8$ pts) Use the first derivative test to find and classify all local extrema in the interval $(-4,5)$ for the function $\newline$ $f(x)=(5 x-4) e^{4 x}$ $\newline$ If there is more than one local maximum or minimum, list them all. If a local maximum or a local minimum does not occur on the interval, note that in your work. You may assume that $e^{4 x}$ is a positive number for all values of $x$ .

View step-by-step help

Home

Math Problems

Algebra 2

Power rule

Full solution

Q. Calculus

1

: Exam

2

\newline

Name:

\newline

15

. (

8

pts) Use the first derivative test to find and classify all local extrema in the interval

(-4,5)

for the function

\newline

f(x)=(5 x-4) e^{4 x}

\newline

If there is more than one local maximum or minimum, list them all. If a local maximum or a local minimum does not occur on the interval, note that in your work. You may assume that

e^{4 x}

is a positive number for all values of

x

Find First Derivative: To find the local extrema, we first need to find the first derivative of the function $f(x) = (5x - 4)e^{4x}$ . $\newline$ Using the product rule for differentiation, which states that $(uv)' = u'v + uv'$ , where $u = (5x - 4)$ and $v = e^{4x}$ , we get: $\newline$ $f'(x) = (5x - 4)'e^{4x} + (5x - 4)(e^{4x})'$ .
Calculate Derivatives of u and v: Now we calculate the derivatives of u and v. The derivative of $u = (5x - 4)$ is $u' = 5$ , and the derivative of $v = e^{(4x)}$ is $v' = 4e^{(4x)}$ because the derivative of $e^{(ax)}$ is $a\cdot e^{(ax)}$ . $\newline$ So, $f'(x) = 5e^{(4x)} + (5x - 4)(4e^{(4x)})$ .
Simplify First Derivative: Simplify the expression for the first derivative: $\newline$ $f'(x) = 5e^{4x} + 20xe^{4x} - 16e^{4x}.$ $\newline$ Combine like terms: $\newline$ $f'(x) = (5e^{4x} - 16e^{4x}) + 20xe^{4x}.$ $\newline$ $f'(x) = -11e^{4x} + 20xe^{4x}.$
Find Critical Points: To find the critical points, we set the first derivative equal to zero and solve for $x$ : $-11e^{4x} + 20xe^{4x} = 0$ . Factor out $e^{4x}$ since it is always positive and will not affect the sign of the expression: $e^{4x}(-11 + 20x) = 0$ .
Solve for x: Since $e^{4x}$ is never zero, we only need to set the remaining factor equal to zero: $\newline$ $-11 + 20x = 0$ . $\newline$ Solve for x: $\newline$ $20x = 11$ . $\newline$ x = $\frac{11}{20}$ .
Use First Derivative Test: We have found a critical point at $x = \frac{11}{20}$ . Now we need to use the first derivative test to classify this critical point as a local maximum, minimum, or neither. $\newline$ We do this by choosing test points on either side of $x = \frac{11}{20}$ and plugging them into the first derivative to check the sign.
Test Point to the Left: Choose a test point to the left of $x = \frac{11}{20}$ , for example, $x = 0$ . Plug $x = 0$ into $f'(x)$ : $f'(0) = -11e^{4\cdot 0} + 20\cdot 0\cdot e^{4\cdot 0} = -11e^0 = -11$ .Since $e^0 = 1$ , $f'(0) = -11$ , which is negative. This means the function is decreasing to the left of $x = \frac{11}{20}$ .
Test Point to the Right: Choose a test point to the right of $x = \frac{11}{20}$ , for example, $x = 1$ . Plug $x = 1$ into $f'(x)$ : $f'(1) = -11e^{(4\cdot 1)} + 20\cdot 1\cdot e^{(4\cdot 1)} = -11e^4 + 20e^4 = 9e^4$ .Since $e^4$ is positive, $f'(1) = 9e^4$ , which is positive. This means the function is increasing to the right of $x = \frac{11}{20}$ .
Determine Local Minimum: Since the function changes from decreasing to increasing at $x = \frac{11}{20}$ , we have a local minimum at this point. $\newline$ Now we need to check if this local minimum occurs within the interval $(-4, 5)$ .
Check Interval: The critical point $x = \frac{11}{20}$ is approximately $0.55$ , which is within the interval $(-4, 5)$ . Therefore, we have a local minimum at $x = \frac{11}{20}$ within the given interval.
Evaluate $f(x)$ : Finally, we need to evaluate $f(x)$ at $x = \frac{11}{20}$ to find the value of the local minimum. $f\left(\frac{11}{20}\right) = \left(5\cdot\left(\frac{11}{20}\right) - 4\right)e^{4\cdot\left(\frac{11}{20}\right)}.$
Simplify Expression: Simplify the expression: $\newline$ $f(\frac{11}{20}) = (\frac{11}{4} - 4)e^{\frac{44}{20}}$ . $\newline$ $f(\frac{11}{20}) = (\frac{11}{4} - \frac{16}{4})e^{\frac{44}{20}}$ . $\newline$ $f(\frac{11}{20}) = (-\frac{5}{4})e^{\frac{44}{20}}$ .
Find Local Minimum Value: We have found the local minimum value of the function $f(x)$ at $x = \frac{11}{20}$ , which is $f\left(\frac{11}{20}\right) = \left(-\frac{5}{4}\right)e^{\frac{44}{20}}$ .