Q. 44 - 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.$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\mathbf{(}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathit{x}\mathbf{-}\mathbf{1}\mathbf{)}$

Step-by-Step Solution

Answer

Ans: The local maximum of the function f(x) is $\mathit{x}\mathbf{=}\mathbf{-}\mathbf{2}$

The local minimum of the function f(x) is $\mathit{x}\mathbf{=}\mathbf{1}$

Step 1. Given Information:

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

Step 2. Finding the derivative of the function:

$$\begin{gathered} f\left(x\right)=e^x(x^2-x-1) \\ f\text{'}\left(x\right)=e^x(2x-1)+(x^2-x-1)e^x \\ f\text{'}\left(x\right)=e^x(2x-1+x^2-x-1) \\ f\text{'}\left(x\right)=e^x(x^2+x-2) \\ let, f\text{'}\left(x\right)=0 \\ \therefore e^x(x^2+x-2)=0 \\ \begin{array}{r} x^2+x-2=0\\ x^2+2x-x-2=0\\ x(x+2)-(x+2)=0\\ (x+2)(x-1)=0\end{array} \\ x=-2,1 \\ the critical points are x=-2,1 \end{gathered}$$

Step 3. Substituting the values into the function equation:

$$\begin{gathered} f(-2)=e^{(-2)}({(-2)}^2-(-2)-1) \\ =e^{-2}(4+2-1) \\ =5e^{-2} \\ f\left(1\right)=e^{\left(1\right)}({\left(1\right)}^2-(1)-1) \\ =-e \\ \therefore the local maximum at x=-2;and the local mimimum at x=1 \end{gathered}$$

Step 4. Verifying algebraic answers with graphs :