Q. 41 - Calculus

Question

Use the first derivative test to determine the local extrema of each function in Exercises 39- 50. Then verify your algebraic answers with graphs from a calculator or graphing utility.

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

Step-by-Step Solution

Answer

Ans: The local maximum of the f(x) is $\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}$

Step 1. Given Information:

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

Step 2. Finding the derivative of the function

Rewriting the function by simplifying it,

$$\begin{gathered} f\left(x\right)=\frac{1+\mathrm{x}+{\mathrm{x}}^2}{{\mathrm{x}}^2+\mathrm{x}-2} \\ =\frac{\left(x^2+x-2\right)(1+2x)-\left(1+x+x^2\right)(2x+1)}{{\left(x^2+x-2\right)}^2} \\ =\frac{(1+2x)\left(x^2+x-2-1-x-x^2\right)}{{\left(x^2+x-2\right)}^2} \\ =\frac{-3(1+2x)}{{\left(x^2+x-2\right)}^2} \\ f\text{'}\left(x\right)=\frac{-3(1+2x)}{{\left(x^2+x-2\right)}^2} \\ let, f\text{'}\left(x\right)=0 \\ \frac{-3(1+2x)}{{\left(x^2+x-2\right)}^2}=0 \\ -3(1+2x)=0 \\ 1+2x=0 \\ 2x=-1 \\ x=-\frac{1}{2} \end{gathered}$$

Step 3. Substituting the values into the function equation:

$$\begin{gathered} f\left(-\frac{1}{2}\right)=\frac{1-\frac{1}{2}+{\left(-\frac{1}{2}\right)}^2}{{\left(-\frac{1}{2}\right)}^2-\frac{1}{2}-2} \\ =\frac{\frac{1}{2}+\frac{1}{4}}{\frac{1}{4}-\frac{5}{2}} \\ =\frac{{\displaystyle \frac{3}{4}}}{{\displaystyle \frac{-9}{4}}}=\frac{3}{-9}=\frac{1}{-3}<0 \\ \therefore the local maximum is \frac{-1}{2} \end{gathered}$$

Step 4. Verifying algebraic answers with graphs :