Q. 2 - Calculus

Question

Sign analyses for derivatives: For each function f  that follows, find the derivative $f\text{'}.$ Then determine the intervals on which the derivative $f\text{'}$ is positive and the intervals on which the derivative $f\text{'}$  is negative. Record your answers on a sign chart for $f\text{'},$ with tick-marks only at the x-values where $f\text{'}$  is zero or undefined.

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

Step-by-Step Solution

Answer

The derivative of the given function is positive on the intervals $\left(-\infty ,-\frac{2}{\ln 3}\right) \mathrm{and} \left(0,\infty \right)$ and negative on the interval $\left(-\frac{2}{\ln 3},0\right).$ The derivative of the function is zero at $x=0 \mathrm{and} x=-\frac{2}{\ln 3}.$

Step 1. Given Information.

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

Step 2. Find the derivative.

To find the derivative of the given function, we will use the product rule of differentiation.

So,

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

Now, let's find the critical point by putting the above equation to zero,

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

Thus, the critical points are $x=0 \mathrm{and} x=-\frac{2}{\ln 3}.$

Step 3. Determine the intervals.

The intervals we get by the critical points are $\left(-\infty ,-\frac{2}{\ln 3}\right),\left(-\frac{2}{\ln 3},0\right) \mathrm{and} \left(0,\infty \right).$

Now, let's take the interval $\left(-\infty ,-\frac{2}{\ln 3}\right),$ to determine where the derivative of the function is positive or negative.

$$\begin{gathered} \mathrm{Let} x=-3 \\ f\text{'}\left(-3\right)=\left(-3\right)3^{\left(-3\right)}\left(\left(-3\right)\ln 3+2\right) \\ f\text{'}\left(-3\right)=\frac{-3}{27}\left(-3\ln 3+2\right) \\ f\text{'}\left(-3\right)=0.143 \end{gathered}$$

Since $f\text{'}\left(x\right)>0$ thus the $f\text{'} \mathrm{is} \mathrm{positive} \mathrm{on} \mathrm{the} \mathrm{interval} \left(-\infty ,-\frac{2}{\ln 3}\right).$

Now, the interval $\left(-\frac{2}{\ln 3},0\right),$

$$\begin{gathered} \mathrm{Let} x=-1 \\ f\text{'}\left(-1\right)=\left(-1\right)3^{\left(-1\right)}\left(\left(-1\right)\ln 3+2\right) \\ f\text{'}\left(-1\right)=-\frac{1}{3}\left(-\ln 3+2\right) \\ f\text{'}\left(-1\right)=-0.3 \end{gathered}$$

Since $f\text{'}\left(x\right)<0$ thus the $f\text{'} \mathrm{is} \mathrm{negative} \mathrm{on} \mathrm{the} \mathrm{interval} \left(-\frac{2}{\ln 3},0\right).$

Step 4. Determine the intervals.

Now, the interval $\left(0,\infty \right),$

$$\begin{gathered} \mathrm{Let} x=3 \\ f\text{'}\left(3\right)=\left(3\right)3^{\left(3\right)}\left(\right(3)\ln 3+2) \\ f\text{'}\left(3\right)=81(3\ln 3+2) \\ f\text{'}\left(3\right)=428.96 \end{gathered}$$

Since $f\text{'}\left(x\right)>0,$ thus the $f\text{'} \mathrm{is} \mathrm{poisitve} \mathrm{in} \mathrm{the} \mathrm{interval} \left(0,\infty \right).$