Chapter 1.4, Problem 47E

a.

To determine

To compute: The difference quotient $\frac{f\left(1+h\right)-f\left(1\right)}{h}$ .

Answer to Problem 47E

  $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{e^{1+h}-e}{h}$

Explanation of Solution

Given information:

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

Calculation:

For $f\left(x\right)=e^x$ .

  $\begin{array}{c}\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{e^{1+h}-e^1}{h}\\ =\frac{e^{1+h}-e}{h}\end{array}$

b.

To determine

To estimate: The limit at $h\to 0$ in $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{e^{1+h}-e}{h}$ .

Answer to Problem 47E

2.1718

Explanation of Solution

Given information:

  $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{e^{1+h}-e}{h}$

Calculation:

From part (a) $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{e^{1+h}-e}{h}$ .

Using graphing calculator:

  

From the graph it can be seen that near $h\to 0^{+}\text{}\text{and }h\to 0^{-}$ , the value of y approaches to 2.718.

On calculation:

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0}{\lim }\frac{e^{1+h}-e}{h}\\ \text{Apply l'Hopital rule}\\ \underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0}{\lim }\frac{\frac{d\left(e^{1+h}-e\right)}{dh}}{\frac{d\left(h\right)}{dh}}\\ =\underset{h\to 0}{\lim }\frac{e^{1+h}-0}{1}\\ \text{Apply limit}\\ \underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=e\\ \approx 2.718\end{array}$

c.

To determine

To compare: The value obtained in part (b) to e .

Answer to Problem 47E

The value obtained in part b is almost same as e (given ).

Explanation of Solution

Given information:

  $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{e^{1+h}-e}{h}$ and the limit at $h\to 0$ in $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{e^{1+h}-e}{h}$ is 2.718.

Calculation:

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0}{\lim }\frac{e^{1+h}-e}{h}\\ \text{Apply l'Hopital rule}\\ \underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0}{\lim }\frac{\frac{d\left(e^{1+h}-e\right)}{dh}}{\frac{d\left(h\right)}{dh}}\\ =\underset{h\to 0}{\lim }\frac{e^{1+h}-0}{1}\\ \text{Apply limit}\\ \underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=e\\ \approx 2.718\end{array}$

Hence the estimated value (using graph) is almost same as the exact value given.

d.

To determine

To find: If the function has tangent at $x=1$ or not, if yes then find the slope.

Answer to Problem 47E

Yes, the function has a tangent at $x=1$ and the slope is e .

Explanation of Solution

Given information:

  $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{e^{1+h}-e}{h}$ .

Concept used:

Find the left hand and right limit of $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{e^{1+h}-e}{h}$ , if the values are equal then slope exist at $x=1$ .

Calculation:

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{e^{1+h}-e}{h}\\ \text{Apply l'Hopital rule}\\ \underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{\frac{d\left(e^{1+h}-e\right)}{dh}}{\frac{d\left(h\right)}{dh}}\\ =\underset{h\to 0^{+}}{\lim }\frac{e^{1+h}-0}{1}\\ \text{Apply limit}\\ \underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=e\end{array}$

Now, find left hand limit as:

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{e^{1+h}-e}{h}\\ \text{Apply l'Hopital rule}\\ \underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{\frac{d\left(e^{1+h}-e\right)}{dh}}{\frac{d\left(h\right)}{dh}}\\ =\underset{h\to 0^{-}}{\lim }\frac{e^{1+h}-0}{1}\\ \text{Apply limit}\\ \underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=e\\ \end{array}$

Since both the left- and right-hand limits are equal. So, tangent at $x=1$ exits and will be equal to:

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0}{\lim }\frac{e^{1+h}-e}{h}\\ \text{Apply l'Hopital rule}\\ \underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0}{\lim }\frac{\frac{d\left(e^{1+h}-e\right)}{dh}}{\frac{d\left(h\right)}{dh}}\\ =\underset{h\to 0}{\lim }\frac{e^{1+h}-0}{1}\\ \text{Apply limit}\\ \underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=e\end{array}$

Hence the slope is e .