Q. 31 - Calculus

Question

Use a sign chart for ${\mathit{f}}^{\mathbf{\text{'}}}$ to determine the intervals on which each function $\mathit{f}$ is increasing or decreasing. Then verify your algebraic answers with graphs from a calculator or graphing utility.

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

Step-by-Step Solution

Answer

Ans:

Intervals of the given function are 

Decreasing  at $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{1}\mathbf{)}$

Increasing at $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{\infty }\mathbf{)}$

Step 1. Given information:

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

Step 2. Finding the derivative of the function:

$$\begin{gathered} f\left(x\right)=e^x(x-2) \\ {f}^{\text{'}}\left(x\right)=(x-2)e^x+e^x\left(1\right) \frac{ \partial }{\partial x}uv=v\frac{ \partial }{\partial x}\left(u\right)+u\frac{ \partial }{\partial x}\left(v\right) \\ {f}^{\text{'}}\left(x\right)=x{e}^x-2e^x+e^x =x{e}^x-e^x \\ {f}^{\text{'}}\left(x\right)=e^x(x-1) \\ let f^{\text{'}}\left(x\right)=0 \\ \therefore e^x(x-1)=0 \\ \boxed{{\mathbf{e}}^{\mathbf{x}}\mathbf{=}\mathbf{0}} \\ \boxed{x-1=0 \\ x=1} \\ x=0,1 \end{gathered}$$

Step 3. Inserting the root points on the number line(Sign chart):

After inserting the root values we can find the increasing and decreasing intervals of the given function. 

Intervals of the given function are 

Increasing at $\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{1}\mathbf{)}$

Decreasing at $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{\infty }\mathbf{)}$

Step 4. Verifying algebraic answers with graphs :