Chapter 1.3, Problem 38E

Solution

a.

To Determine: The interval in which the function is increasing or decreasing.

The function is increasing on the interval of $\left(-\infty ,\infty \right)$ .

Given information:

  $q\left(x\right)=e^x+2$

Graph:

  

From the graph, it can be seen that the function is increasing throughout its entire domain, which is $\left(-\infty ,\infty \right)$ .

b.

To Determine: The function is odd, even or neither.

The function is neither even nor odd.

Given information:

  $q\left(x\right)=e^x+2$

Calculation:

A function is even if $f\left(-x\right)=f\left(x\right)$

Check if $f\left(-x\right)=f\left(x\right)$ :

  $e^{-x}+2\ne e^x+2$

Since $e^{-x}+2\ne e^x+2$ , the function is not even.

A function is odd if $f\left(-x\right)=-f\left(x\right)$

Multiply $-1$ by $e^x+2$ :

  $-f\left(x\right)=-\left(e^x+2\right)$

Since $\begin{array}{l}-\left(e^x+2\right)\ne e^x+2\\ -e^x-2\ne e^x+2\end{array}$ , the function is not odd.

So, the function is neither even nor odd.

c.

To Determine: The extrema of the function, if any.

There are no extrema points.

Given information:

  $q\left(x\right)=e^x+2$

Calculation:

Find the first derivative of the function.

  $e^x$

Differentiate using the Exponential Rule which states that $\frac{d}{dx}\left[a^x\right]$ is $a^x\ln \left(a\right)$ where $a=e$ .

  $f^{\prime \prime }\left(x\right)=e^x$

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve:

  $e^x=0$

Since there is no value of $x$ that makes the first derivative equal to $0$ , there are no local extrema.

Thus, there is no extrema for the function $q\left(x\right)=e^x+2$ .

d.

To Determine: The graph of the function related to a graph of one of the twelve basic functions.

The graph is related to exponential function.

Given information:

  $q\left(x\right)=e^x+2$

Calculation:

  

From the above graph, it can be seen that the function $q\left(x\right)=e^x+2$ is a exponential function since, it shifted $2$ units up.