let (f(x)=x^{2}e^{15x}). (a) find the critical values of (f(x)). if there are none, enter none. if there are…

Explanation:

Step1: Find the derivative of $f(x)$

Use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = x^{2}$ and $v = e^{15x}$. $u^\prime=2x$ and $v^\prime = 15e^{15x}$. So $f^\prime(x)=2x e^{15x}+15x^{2}e^{15x}=x e^{15x}(2 + 15x)$.

Step2: Find the critical values

Set $f^\prime(x)=0$. Since $e^{15x}>0$ for all real $x$, we solve $x(2 + 15x)=0$. $x = 0$ or $2+15x=0$, which gives $x = 0$ or $x=-\frac{2}{15}$.

Step3: Determine the sign of $f^\prime(x)$ on intervals

The critical values divide the real - line into intervals: $\left(-\infty,-\frac{2}{15}\right)$, $\left(-\frac{2}{15},0\right)$ and $(0,\infty)$. Choose test points: $x=-1$ for $\left(-\infty,-\frac{2}{15}\right)$, $x =-\frac{1}{15}$ for $\left(-\frac{2}{15},0\right)$ and $x = 1$ for $(0,\infty)$. For $x=-1$, $f^\prime(-1)=(-1)e^{-15}(2-15)=13e^{-15}>0$. For $x =-\frac{1}{15}$, $f^\prime\left(-\frac{1}{15}\right)=\left(-\frac{1}{15}\right)e^{-1}(2 - 1)=-\frac{1}{15e}<0$. For $x = 1$, $f^\prime(1)=1\times e^{15}(2 + 15)=17e^{15}>0$. So $f(x)$ is increasing on $\left(-\infty,-\frac{2}{15}\right)\cup(0,\infty)$ and decreasing on $\left(-\frac{2}{15},0\right)$.

Step4: Find relative extrema

Since $f(x)$ changes from increasing to decreasing at $x =-\frac{2}{15}$, $f\left(-\frac{2}{15}\right)=\left(-\frac{2}{15}\right)^{2}e^{-2}=\frac{4}{225e^{2}}$ is a relative maximum. Since $f(x)$ changes from decreasing to increasing at $x = 0$, $f(0)=0$ is a relative minimum.

Answer:

(a) $x = 0,-\frac{2}{15}$ (b) $\left(-\infty,-\frac{2}{15}\right)\cup(0,\infty)$ (c) $\left(-\frac{2}{15},0\right)$ (d) $x = 0$ (e) $x=-\frac{2}{15}$