Q. 31 - 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.) 

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

Step-by-Step Solution

Answer

The function has a local maximum at $x=-\frac{1}{2}$.

Step 1. Given information.

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

Step 2. Critical points.

On differentiating the given function, we get,

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

The critical point is when $f\text{'}$ that is at $x=-\frac{1}{2}$

Step 3. Second Derivative Test.

Again differentiating, we get,

$$\begin{gathered} f^{\text{'}\text{'}}\left(x\right)=\frac{d}{dx}\frac{-3(2x+1)}{{\left(x^2+x-2\right)}^2} \\ =\frac{18\left(x^2+x+1\right)}{{\left(x^2+x-2\right)}^3} \\ {f}^{\text{'}\text{'}}\left(-\frac{1}{2}\right)=\frac{18\left({\left(-\frac{1}{2}\right)}^2+\left(-\frac{1}{2}\right)+1\right)}{{\left({\left(-\frac{1}{2}\right)}^2+\left(-\frac{1}{2}\right)-2\right)}^3} \\ =\frac{18\left(\frac{1}{4}-\frac{1}{2}+1\right)}{{\left(\frac{1}{4}-\frac{1}{2}-2\right)}^3} \\ =-\frac{32}{17}<0 \\ Therefore, \\ x=-\frac{1}{2} \end{gathered}$$

shows the point where the function has local maximum.

Step 4. Verification.

The graph of the function is ,

which verifies the local maximum at $x=-\frac{1}{2}.$