Consider the following function. Use a graphing utility to confirm your answers for parts (a) through ©. (If an answer does not exist, enter
DNE.)

(a) Find the critical numbers of . (Enter your answers as a comma-separated list.)
(b) Find the open intervals on which the function is increasing or decreasing. (Enter your answers using interval notation.)
increasing
decreasing
© Apply the First Derivative Test to identify all relative extrema.
relative maximum

Ask by Cox Sullivan. in the United States

Mar 31,2025

Question

Consider the following function. Use a graphing utility to confirm your answers for parts (a) through ©. (If an answer does not exist, enter
DNE.)

(a) Find the critical numbers of . (Enter your answers as a comma-separated list.)
(b) Find the open intervals on which the function is increasing or decreasing. (Enter your answers using interval notation.)
increasing
decreasing
© Apply the First Derivative Test to identify all relative extrema.
relative maximum

Ask by Cox Sullivan. in the United States

Mar 31,2025

UpStudy AI · Accepted Answer

To find the critical numbers of the function $$ f(x) = 13x + \frac{1}{x} $$, we first need to compute the derivative $$ f'(x) $$ and set it equal to zero.
Calculate the derivative of $$ f(x) $$:
$$
f'(x) = 13 - \frac{1}{x^2}
$$
Now, set the derivative equal to zero to find critical points:
$$
13 - \frac{1}{x^2} = 0
$$
Solving for $$ x $$, we rearrange the equation:
$$
\frac{1}{x^2} = 13 \implies x^2 = \frac{1}{13} \implies x = \pm \frac{1}{\sqrt{13}}
$$
Thus, the critical numbers are $$ x = \frac{1}{\sqrt{13}}, -\frac{1}{\sqrt{13}} $$.
Next, we determine the intervals of increase and decrease by analyzing the sign of $$ f'(x) $$. We will test intervals around the critical points $$ x = -\frac{1}{\sqrt{13}} $$ and $$ x = \frac{1}{\sqrt{13}} $$.
Choose test points in the intervals:
1. $$ (-\infty, -\frac{1}{\sqrt{13}}) $$
2. $$ (-\frac{1}{\sqrt{13}}, \frac{1}{\sqrt{13}}) $$
3. $$ (\frac{1}{\sqrt{13}}, \infty) $$

Evaluate $$ f'(x) $$ at these points:
- For $$ x = -1 $$ (in the first interval):
$$
f'(-1) = 13 - \frac{1}{(-1)^2} = 13 - 1 = 12 > 0 \quad \text{(increasing)}
$$
- For $$ x = 0 $$ (in the second interval, note $$ f'(x) $$ is undefined at $$ x = 0 $$):
- For $$ x = 1 $$ (in the third interval):
$$
f'(1) = 13 - \frac{1}{1^2} = 13 - 1 = 12 > 0 \quad \text{(increasing)}
$$

Thus, the function is:
- Increasing on $$ (-\infty, -\frac{1}{\sqrt{13}}) $$ and $$ (\frac{1}{\sqrt{13}}, \infty) $$
- Decreasing on $$ (-\frac{1}{\sqrt{13}}, \frac{1}{\sqrt{13}}) $$.
Now, we apply the First Derivative Test to identify relative extrema. Since $$ f'(x) $$ changes from positive to negative at $$ x = -\frac{1}{\sqrt{13}} $$, there is a relative maximum at this point. We will calculate $$ f(-\frac{1}{\sqrt{13}}) $$ to find the corresponding $$ y $$-value.
Calculate $$ f(-\frac{1}{\sqrt{13}}) $$:
$$
f\left(-\frac{1}{\sqrt{13}}\right) = 13\left(-\frac{1}{\sqrt{13}}\right) + \frac{1}{-\frac{1}{\sqrt{13}}} = -\frac{13}{\sqrt{13}} - \sqrt{13} = -\sqrt{13} - \sqrt{13} = -2\sqrt{13}
$$
Thus, the relative maximum is at $$ (x, y) = \left(-\frac{1}{\sqrt{13}}, -2\sqrt{13}\right) $$.
(a) Critical numbers: $$ \frac{1}{\sqrt{13}}, -\frac{1}{\sqrt{13}} $$ (b) Increasing: $$ (-\infty, -\frac{1}{\sqrt{13}}) \cup (\frac{1}{\sqrt{13}}, \infty) $$ Decreasing: $$ (-\frac{1}{\sqrt{13}}, \frac{1}{\sqrt{13}}) $$ (c) Relative maximum: $$ (x, y) = \left(-\frac{1}{\sqrt{13}}, -2\sqrt{13}\right) $$

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