Chapter 4.8, Problem 40E

a.

To determine

To graph: The equation $f\left(x\right)=e^x-2x$ .

Explanation of Solution

Given information:

The equation $f\left(x\right)=e^x-2x$ .

Graph:

The graph of the equation $f\left(x\right)=e^x-2x$ is provided below,

  

Interpretation:

The graph of equation $f\left(x\right)=e^x-2x$ is always positive so whole lie on the upper side.

b.

To determine

To calculate: The antiderivative of the equation $f\left(x\right)=e^x-2x$ and sketch the graph of antiderivative $F$ with condition $F\left(0\right)=1$ .

Answer to Problem 40E

The antiderivative of the equation $f\left(x\right)=e^x-2x$ is $F\left(x\right)=e^x-x^2$ and the graph of antiderivative is provided below,

  

Explanation of Solution

Given information:

The equation is given as:

  $f\left(x\right)=e^x-2x$ and condition $F\left(0\right)=1$ .

Formula used:

The function F is antiderivative of $f\left(x\right)$ on an interval $I$ if $F^{`}\left(x\right)=f\left(x\right)$ for all $x$ in $I$

Power rule:

  $\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n$

Calculation:

Consider the equation,

  $f\left(x\right)=e^x-2x$

The antiderivative of $f\left(x\right)$ is,

  $F^{`}\left(x\right)=f\left(x\right)$

So, by reverse power rule Integrating both sides of equation,

  $\begin{array}{c}{\displaystyle \int F^{`}\left(x\right)}={\displaystyle \int f\left(x\right)}\\ F\left(x\right)={\displaystyle \int f\left(x\right)}\end{array}$

Therefore, integrating the function

  $\begin{array}{c}f\left(x\right)=e^x-2x\\ ={\displaystyle \int e^x}dx-2{\displaystyle \int x}dx\\ =e^x-2\frac{x^2}{2}+C\\ =e^x-x^2+C\end{array}$

Thus, the of the equation is $F\left(x\right)=e^x-x^2+C$ .

Now applying the condition $F\left(0\right)=1$ :

  $\begin{array}{c}F\left(0\right)=e^0-0^2+C\\ 1=1+C\\ C=0\end{array}$

Put the value of $C$ in the equation,

  $F\left(x\right)=e^x-x^2$

So, now using the graph device and the graph of antiderivative $F\left(x\right)=e^x-x^2$ which satisfy thecondition $F\left(0\right)=1$ is provided below,

  

c.

To determine

To calculate: The antiderivative of the equation $f\left(x\right)=e^x-2x$ .

Answer to Problem 40E

The antiderivative of the equation $f\left(x\right)=e^x-2x$ is $F\left(x\right)=e^x-x^2+C$ .

Explanation of Solution

Given information:

The equation is given as:

  $f\left(x\right)=e^x-2x$ .

Formula used:

The function F is antiderivative of $f\left(x\right)$ on an interval $I$ if $F^{`}\left(x\right)=f\left(x\right)$ for all $x$ in $I$

Power rule:

  $\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right)=x^n$

Calculation:

Consider the equation,

  $f\left(x\right)=e^x-2x$

The antiderivative of $f\left(x\right)$ is,

  $F^{`}\left(x\right)=f\left(x\right)$

So, by reverse power rule Integrating both sides of equation,

  $\begin{array}{c}{\displaystyle \int F^{`}\left(x\right)}={\displaystyle \int f\left(x\right)}\\ F\left(x\right)={\displaystyle \int f\left(x\right)}\end{array}$

Therefore, integrating the function

  $\begin{array}{c}f\left(x\right)=e^x-2x\\ ={\displaystyle \int e^x}dx-2{\displaystyle \int x}dx\\ =e^x-2\frac{x^2}{2}+C\\ =e^x-x^2+C\end{array}$

Thus, the of the equation is $F\left(x\right)=e^x-x^2+C$ .

d.

To determine

To graph: The equation $F\left(x\right)=e^x-x^2+C$ and compare with the graph of equation $F\left(x\right)=e^x-x^2$ .

Explanation of Solution

Given information:

The equation $F\left(x\right)=e^x-x^2+C$ .

Graph:

The graph of the equation $F\left(x\right)=e^x-x^2+C$ is provided below,

  

The graph of equation $F\left(x\right)=e^x-x^2$ is provided below,

  

Interpretation:

The graphs of equation $F\left(x\right)=e^x-x^2+C$ is define for different different values of $C$ andwhen the value of $C$ is 3.

Then the graph of equation $F\left(x\right)=e^x-x^2+C$ is same as equation $F\left(x\right)=e^x-x^2$ .