let ( f(x)=e^{-3x^{2}}).\na. compute ( f(x)).\n( f(x)=)\n\nb. find the ( x - value(s)) where ( f(x) = 0)…

Explanation:

Step1: Find the first - derivative using chain - rule

Let $u=-3x^{2}$, then $f(x)=e^{u}$. The chain - rule states that $\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}$. Since $\frac{df}{du}=e^{u}$ and $\frac{du}{dx}=-6x$, we have $f^{\prime}(x)=e^{-3x^{2}}\cdot(-6x)=-6xe^{-3x^{2}}$.

Step2: Find the second - derivative using product - rule

The product - rule states that if $y = uv$, then $y^{\prime}=u^{\prime}v + uv^{\prime}$, where $u=-6x$ and $v = e^{-3x^{2}}$. We know that $u^{\prime}=-6$ and $v^{\prime}=e^{-3x^{2}}\cdot(-6x)$. So $f^{\prime\prime}(x)=(-6)e^{-3x^{2}}+(-6x)\cdot(-6x)e^{-3x^{2}}=e^{-3x^{2}}(36x^{2}-6)$.

Step3: Solve $f^{\prime\prime}(x) = 0$

Set $f^{\prime\prime}(x)=e^{-3x^{2}}(36x^{2}-6)=0$. Since $e^{-3x^{2}}\gt0$ for all real $x$, we solve $36x^{2}-6 = 0$. Rearranging gives $36x^{2}=6$, then $x^{2}=\frac{6}{36}=\frac{1}{6}$, so $x=\pm\frac{1}{\sqrt{6}}$.

Answer:

a. $f^{\prime\prime}(x)=e^{-3x^{2}}(36x^{2}-6)$ b. $x =-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}$