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Suppose that $f(x)=(1+8 x) e^{5 x}$.


AFind all critical values of $f$. (If there are multiple values, enter them separated by commas. If there are no critical points, enter DNE.)

$x=$


BUse interval notation to indicate where $f$ is increasing. (If there are multiple intervals, enter them separated by commas.)


CUse interval notation to indicate where $f$ is decreasing. (If there are multiple intervals, enter them separated by commas.)


DList the $x$-coordinates of all local maxima of $f$. (If there are multiple values, enter them separated by commas. If there are no local maxima, enter DNE.)

$x=$


EList the $x$-coordinates of all local minima of $f$. (If there are multiple values, enter them separated by commas. If there are no local minima, enter DNE.)

$x=$

Verified Answer :

  • Calculate the derivative \(f'(x) = e^{5x}(13 + 40x)\) to find the critical point \(x = -\frac{13}{40}\).
  • Evaluate the sign of \(f'(x)\) to determine that the function is decreasing on \((-\infty, -\frac{13}{40})\) and increasing on \((-\frac{13}{40}, \infty)\).
  • Apply the first derivative test at \(x = -\frac{13}{40}\) to conclude that the function has a local minimum at this point and no local maxima.
  • Final answers:
A\(x = -\frac{13}{40}\)
B\((-\frac{13}{40}, \infty)\)
C\((-\infty, -\frac{13}{40})\)
D\(\text{DNE}\)
E\(x = -\frac{13}{40}\)

Explanation

  1. Finding Critical Values

To analyze the function \(f(x) = (1 + 8x)e^{5x}\), we first determine its domain, which is all real numbers \((-\infty, \infty)\). Critical values occur where the first derivative \(f'(x)\) is zero or undefined. We apply the product rule, \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\), with \(u(x) = 1 + 8x\) and \(v(x) = e^{5x}\). The derivative of \(u(x)\) is \(u'(x) = 8\), and the derivative of \(v(x)\) is \(v'(x) = 5e^{5x}\). Combining these, we get:
\[f'(x) = 8e^{5x} + (1 + 8x)(5e^{5x})\]
Factoring out \(e^{5x}\), we have:
\[f'(x) = e^{5x}(8 + 5(1 + 8x)) = e^{5x}(8 + 5 + 40x) = e^{5x}(13 + 40x)\]
Since \(e^{5x}\) is never zero, \(f'(x) = 0\) when \(13 + 40x = 0\), which gives the critical value \(x = -\frac{13}{40}\) or \(-0.325\).

  1. Intervals of Increase and Decrease

To determine the intervals of increase and decrease, we analyze the sign of \(f'(x) = e^{5x}(13 + 40x)\). Since \(e^{5x} > 0\) for all real \(x\), the sign of \(f'(x)\) is determined solely by the factor \((13 + 40x)\).

  • For \(x < -\frac{13}{40}\), we can test a point such as \(x = -1\). Then \(13 + 40(-1) = -27 < 0\), so \(f'(x) < 0\), and the function is decreasing.
  • For \(x > -\frac{13}{40}\), we can test a point such as \(x = 0\). Then \(13 + 40(0) = 13 > 0\), so \(f'(x) > 0\), and the function is increasing.

Thus, \(f\) is decreasing on \((-\infty, -\frac{13}{40})\) and increasing on \((-\frac{13}{40}, \infty)\).

  1. Local Extrema

We use the First Derivative Test to classify the critical point \(x = -\frac{13}{40}\).

  • Since the sign of \(f'(x)\) changes from negative to positive at \(x = -\frac{13}{40}\), the function transitions from decreasing to increasing.
  • Therefore, the function has a local minimum at \(x = -\frac{13}{40}\). There are no local maxima.
  1. Summary of Results

The critical value is \(x = -\frac{13}{40}\). The function is increasing on \((-\frac{13}{40}, \infty)\) and decreasing on \((-\infty, -\frac{13}{40})\). There is a local minimum at \(x = -\frac{13}{40}\) and no local maxima.
Final answers:

ACritical value: \(x = -\frac{13}{40}\)
BIncreasing on: \((-\frac{13}{40}, \infty)\)
CDecreasing on: \((-\infty, -\frac{13}{40})\)
DLocal maxima: DNE
ELocal minima: \(x = -\frac{13}{40}\)

Examples

In fields like economics or physics, functions like \(f(x) = (1+8x)e^{5x}\) are used to model growth or decay processes. Finding critical points allows us to identify the "turning points" of a system—for example, determining the time at which a company's profit stops falling and starts rising or when a chemical reaction reaches its minimum concentration before stabilization. By calculating the derivative and identifying where it is zero, you can optimize processes and predict when significant changes in a trend will occur, providing a foundation for data-driven decision making.

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