Q. 39 - 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.

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

Step-by-Step Solution

Answer

Ans: The local extrema value of f(x) are $x=\left\{0,4\right\}$

Step 1. Given Information

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

Step 2. Finding the derivative of the function

Rewriting the function by simplifying it,

$$\begin{gathered} f\left(x\right)={(x-2)}^2(x+1) \\ f\left(x\right)=(x^2-4x+4)(1+x) \\ f\left(x\right)=x^2+x^3-4x-4x^2+4+4x \\ f\left(x\right)=x^3-3x^2+4 \\ f\text{'}\left(x\right)=3x^2-6x \\ lets, f\text{'}\left(x\right)=0 \\ \therefore 3x^2-6x=0 \\ 3x(x-2)=0 \\ x=\{0,2\} \\ substitute x=-1,1,3 into f\text{'}\left(x\right)=3x^2-6x \\ f\text{'}(-1)=3{(-1)}^2-6(-1) \\ =9>0 \\ f\text{'}\left(1\right)=3{\left(1\right)}^2-6\left(1\right) \\ =-3<0 \\ f\text{'}\left(3\right)=3{\left(3\right)}^2-6\left(3\right) \\ =9>0 \end{gathered}$$

Step 3. Finding local extrema of the function on the number line(Sign chart):

$$\begin{gathered} From the graph identify the local minimum at x=2 indicated on the chart \\ with the first derivative test, \sin ce f\text{'}\left(x\right) changes sign from negative to \\ positive at x=2. \\ f\left(0\right)=(0-2{)}^2(1+0) \\ =4 (local maximum\left) \\ f\right(2)=(2-2{)}^2(1+2) \\ =0 (local minimum) \\ \therefore the extreme values of the function are x=\{0,4\} \end{gathered}$$

Step 4. Verifying algebraic answers with graphs :