Chapter 3.1, Problem 2E

(a)

To determine

To sketch: The graph of the function and obtain the point when the function crosses the y-axis.

Answer to Problem 2E

The graph of the function $f\left(x\right)=e^x$ is shown below in Figure 1.

The function $f\left(x\right)=e^x$ crosses the y axis at $\left(0,1\right)$.

Explanation of Solution

Given:

The function is $f\left(x\right)=e^x$.

Formula used: Derivative of Exponential Function

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

Calculation:

Obtain the derivative of $f\left(x\right)=e^x$.

Use the derivative of exponential function in equation (1).

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

Since the given function and its derivative are same, $f\left(x\right)=f^{\prime }\left(x\right)$.

The graph of the function $f\left(x\right)=e^x$ is shown below in Figure 1.

From the graph, it is observed that the function $f\left(x\right)=e^x$ increases for all x, because the derivative of the function $f^{\prime }\left(x\right)=e^x$ is positive for all x.

Moreover, the function is closer to zero as x approaches minus infinity and it is closer to infinity as x approaches plus infinity.

That is, $\underset{x\to -\infty }{\lim }{e}^x=0\text{ and }\underset{x\to \infty }{\lim }{e}^x=\infty$.

Substitute 0 for x in $f\left(x\right)=e^x$,

$\begin{array}{c}f\left(0\right)=e^0\\ =1\end{array}$

Therefore, the function is crosses the y axis at $\left(0,1\right)$.

(b)

To determine

To describe: The type of functions $f\left(x\right)$ and $g\left(x\right)$ and compare the differentiation formulas for $f\left(x\right)$ and $g\left(x\right)$.

Answer to Problem 2E

Both the functions $f\left(x\right)$ and $g\left(x\right)$ are increasing function, and $f\left(x\right)$ is an exponential function.

The differentiation formulas for $f\left(x\right)\text{ and }g\left(x\right)$ are $f^{\prime }\left(x\right)=e^x\text{ and }{g}^{\prime }\left(x\right)=e{x}^{e-1}$.

Explanation of Solution

Given:

The function are $f\left(x\right)=e^x$ and $g\left(x\right)=x^e$.

Formula used: Power Rule

If n is a real number, then $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ (2)

Calculation:

The graph of the function $g\left(x\right)=x^e$ is shown below in Figure 2.

From the graph, it is observed that the function $g\left(x\right)=x^e$ is defined for only positive value of x and $g^{\prime }\left(x\right)=e{x}^{e-1}$ is positive for all x.

Thus, $g\left(x\right)=x^e$ is an increasing function for all value of x.

From part (a), $f\left(x\right)=e^x$ is an exponential function which is defined for all value of x and it is an increasing function.

The derivatives of both the functions $f^{\prime }\left(x\right)=e^x$ and $g^{\prime }\left(x\right)=e{x}^{e-1}$ are increasing when x increases.

Therefore, both the functions $f\left(x\right)$ and $g\left(x\right)$ are increasing functions and $f\left(x\right)$ is an exponential function.

Obtain the derivatives of $f\left(x\right)\text{ and }g\left(x\right)$.

From part (a), the derivative of $f\left(x\right)$ is $f^{\prime }\left(x\right)=e^x$.

Since the derivative of $g\left(x\right)$ is $g^{\prime }\left(x\right)$, $g^{\prime }\left(x\right)=\frac{d}{dx}\left(x^e\right)$.

Apply the Power rule (2), $g^{\prime }\left(x\right)=e{x}^{e-1}$.

Thus, the derivative of $g\left(x\right)$ is $g^{\prime }\left(x\right)=e{x}^{e-1}$.

Therefore, the differentiation formulas for $f\left(x\right)$ and $g\left(x\right)$ are $f^{\prime }\left(x\right)=e^x\text{ and }{g}^{\prime }\left(x\right)=e{x}^{e-1}$.

(c)

To determine

To identify: The function which grows more rapidly when x is large.

Answer to Problem 2E

The function $f\left(x\right)=e^x$ is faster rate compared to $g\left(x\right)=x^e$ when x is large.

Explanation of Solution

Given:

The function are $f\left(x\right)=e^x$ and $g\left(x\right)=x^e$.

The graph of the functions $f\left(x\right)=e^x$ and $g\left(x\right)=x^e$ is shown below in Figure 3.

From the graph, it is observed that the value of $f\left(x\right)=e^x$ is larger than $g\left(x\right)=x^e$ when $x=1$.

That is, $f\left(1\right)>g\left(1\right)$.

Therefore, the function $f\left(x\right)=e^x$ grows more rapidly than $g\left(x\right)=x^e$ when x is large.