Question

use the first derivative test to find the relative extrema of (f(x)=\frac{x - 6}{e^{x}}). (f) has a select an answer at (x=)

Explanation:

Step1: Find the derivative using quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x - 6$, $u^\prime=1$, $v = e^{x}$, and $v^\prime=e^{x}$. So, $f^\prime(x)=\frac{1\cdot e^{x}-(x - 6)e^{x}}{(e^{x})^{2}}=\frac{e^{x}-(x - 6)e^{x}}{e^{2x}}=\frac{e^{x}(1-(x - 6))}{e^{2x}}=\frac{7 - x}{e^{x}}$.

Step2: Find the critical points

Set $f^\prime(x)=0$. Since $e^{x}
eq0$ for all real $x$, we solve $7 - x = 0$. So, $x = 7$ is the critical point.

Step3: Use the first - derivative test

Choose test points. Let's choose $x = 6$ and $x = 8$.
For $x = 6$, $f^\prime(6)=\frac{7-6}{e^{6}}=\frac{1}{e^{6}}>0$.
For $x = 8$, $f^\prime(8)=\frac{7 - 8}{e^{8}}=-\frac{1}{e^{8}}<0$.
Since $f^\prime(x)$ changes sign from positive to negative at $x = 7$, $f(x)$ has a relative maximum at $x = 7$.

Answer:

Relative maximum; $7$