Q. 40 - 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.

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

Step-by-Step Solution

Answer

Ans: The local extrema value of f(x) are 

Max at $\mathit{x}\mathbf{=}\mathbf{0}$

Min at $\mathit{x}\mathbf{=}\mathbf{\pm }\sqrt{\frac{\mathbf{1}}{\mathbf{2}}}$

Step 1. Given Information

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

Step 2. Finding the derivative of the function

Rewriting the function by simplifying it,

$$\begin{gathered} f\left(x\right)=x^2(x-1)(x+1) \\ =x^2(x^2-1) \\ =x^4-x^2 \\ f\text{'}\left(x\right)=4x^3-2x \\ let, f\text{'}\left(x\right)=0 \\ 4x^3-2x=0 \\ 2x(2x^2-1)=0 \\ \boxed{2x=0 \\ x=0} \\ \boxed{2x^2-1=0 \\ 2x^2=1 \\ x=\pm \sqrt{\frac{1}{2}}} \\ x=0,\pm \sqrt{\frac{1}{2}} \end{gathered}$$

Step 3. Substituting the values into the function equation:

$$\begin{gathered} x=-2,\frac{1}{2},2 into the f\text{'}\left(x\right)=x^4-x^2 \\ f\text{'}(-2)={(-2)}^4-{(-2)}^2 \\ =12>0 \\ f\text{'}\left(2\right)={\left(2\right)}^4-{\left(2\right)}^2 \\ =12>0 \\ f\text{'}\left(\frac{1}{2}\right)={\left(\frac{1}{2}\right)}^4-{\left(\frac{1}{2}\right)}^2 \\ =\frac{-3}{16}<0 \end{gathered}$$

Step 4. Finding local extrema of the function on the number line(Sign chart):

Step 5. Verifying algebraic answers with graphs :