Q. 32 - 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.

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

Step-by-Step Solution

Answer

Ans: It has no solution hence this is a constant function.

Intervals of the given function are 

Increasing at $\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{\infty }\mathbf{)}$

Step 1. Given information:

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

Step 2. Finding the derivative of the function:

$$\begin{gathered} f\left(x\right)=\frac{e^x}{1+e^x} \\ {f}^{\text{'}}\left(x\right)=\frac{(1+e^x)e^x-e^x\left(e^x\right) }{{(1+e^x)}^2} \frac{ \partial }{\partial x}\frac{u}{v}=\frac{v\frac{ \partial }{\partial x}\left(u\right)-u\frac{ \partial }{\partial x}\left(v\right)}{v^2} \\ {f}^{\text{'}}\left(x\right)=\frac{e^x+e^{2x}-e^{2x}}{{(1+e^x)}^2} =\frac{e^x}{{(1+e^x)}^2} \\ {f}^{\text{'}}\left(x\right)=\frac{e^x}{{(1+e^x)}^2} \\ let f^{\text{'}}\left(x\right)=0 \\ \therefore \frac{e^x}{{(1+e^x)}^2}=0 \\ {e}^x={(1+e^x)}^2\times 0 \\ {e}^x=0 \\ \boxed{{\mathbf{e}}^{\mathbf{x}}\mathbf{=}\mathbf{0}} \\ x=no solutions \end{gathered}$$

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

It has no solution hence this is a constant function.

so according to the graph, we can say that the function increases from 1 

Therefore $\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{\infty }\mathbf{)}$

Step 4. Verifying algebraic answers with graphs :