Q. 32 - Calculus

Question

Use the second-derivative test to determine the local extrema of each function $f$ in Exercises $29-40$. If the second-derivative test fails, you may use the first-derivative test. Then verify your algebraic answers with graphs from a calculator or graphing utility. (Note: These are the same functions that you examined with the first-derivative test in Exercises 39–50 of Section 3.2.)

$f\left(x\right)=\frac{(x-1{)}^2}{x+2}$

Step-by-Step Solution

Answer

The local maximum is at $x=-5$ and local minimum is at $x=1.$

Step 1. Given Information.

The given function is $f\left(x\right)=\frac{(x-1{)}^2}{x+2}.$

Step 2. Critical points.

On differentiating the given information, we get,

$$\begin{gathered} f\left(x\right)=\frac{(x-1{)}^2}{x+2} \\ {f}^{\text{'}}\left(x\right)=\frac{d}{dx}\frac{(x-1{)}^2}{x+2} \\ =\frac{\left(\frac{d}{dx}(x-1{)}^2\right)(x+2)-(x-1{)}^2\left(\frac{d}{dx}(x+2)\right)}{(x+2{)}^2} \\ =\frac{\left(2\right(x-1\left)\right)(x+2)-(x-1{)}^2}{(x+2{)}^2} \\ =\frac{(x-1)(x+5)}{(x+2{)}^2} \end{gathered}$$

The critical points are points where $f\text{'}=0$.

Therefore, critical points are $x=1\text{ and }x=-5.$

Step 3. Second-Derivative Test.

Again differentiating the function, we get,

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\frac{(x-1)(x+5)}{(x+2{)}^2} \\ =\frac{18}{(x+2{)}^3} \\ {f}^{\text{'}\text{'}}(-5)=\frac{18}{(-5+2{)}^3} \\ =-\frac{2}{3}<0 \\ {f}^{\text{'}\text{'}}\left(1\right)=\frac{18}{(1+2{)}^3} \\ =\frac{2}{3}>0 \end{gathered}$$

Therefore.

The function has a local minimum at $x=1$ and local maximum at $x=-5.$

Step 4. Verification.

The graph of the function is,

which shows $x=1$ is the point of local minima.