Question

Sign analyses for second derivatives: Repeat the instructions of the previous block of problems, except find sign intervals for the second derivative $f\text{'}\text{'}$instead of the first derivative.

$f\left(x\right)=x^2{3}^x$

Short answer

The second derivative of the given function is the local maximum at $x=-1.8,$and a local minimum at $x=0.$

Step-by-step solution

Step 1. Given Information.

The given function is$f\left(x\right)=x^2{3}^x.$

Step 2. Find the second derivative. 

To find the second derivative, first, we find the first derivative then the second.

So,

$$\begin{gathered} f\left(x\right)=x^2{3}^x \\ f\text{'}\left(x\right)=x^2{3}^x\ln 3+2x{3}^x \\ f\text{'}\text{'}\left(x\right)=x^2\ln 3\left(3^x\ln 3\right)+\left(2x\right)3^x\ln 3+2x\left(3^x\ln 3\right)+2\left(3^x\right) \\ f\text{'}\text{'}\left(x\right)=x^2\ln 3\left(3^x\ln 3\right)+4x\left(3^x\ln 3\right)+2\left(3^x\right) \end{gathered}$$

Step 3. Find sign intervals for the second derivative. 

To find the sign intervals let's find the critical points from the first derivative.

So,

$$\begin{gathered} f\text{'}\left(x\right)=0 \\ {x}^2{3}^x\ln 3+2x{3}^x=0 \\ x{3}^x(x\ln 3+2)=0 \\ x=0\mathrm{and}x=\frac{-2}{\ln 3} \\ x=\frac{-2}{1.09} \\ x=-1.8 \end{gathered}$$

Thus, the critical points are$x=0\text{and}x=-1.8.$

Now, at $x=0,f\text{'}\text{'}\left(x\right)>0$and at$x=-1.8,f\text{'}\text{'}\left(x\right)<0.$