Chapter 1.4, Problem 48E

a.

To determine

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

Answer to Problem 48E

  $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{2\left(2^h-1\right)}{h}$

Explanation of Solution

Given information:

The function is $f\left(x\right)=2^x,\ln 4$ .

Calculation:

For $f\left(x\right)=2^x$ .

  $\begin{array}{c}\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{2^{1+h}-2^1}{h}\\ =\frac{2\left(2^h-1\right)}{h}\end{array}$

b.

To determine

To estimate: The limit at $h\to 0$ in $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{2\left(2^h-1\right)}{h}$ .

Answer to Problem 48E

1.39.

Explanation of Solution

Given information:

  $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{2\left(2^h-1\right)}{h}$

Calculation:

From part (a) $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{2\left(2^h-1\right)}{h}$ .

Using graphing calculator:

  

From the graph it can be seen that for different value of h near $h\to 0^{+}\text{}\text{and }h\to 0^{-}$ , the value of y approaches to 1.39.

On calculation:

c.

To determine

To compare: The value obtained in part (b) to ln(4).

Answer to Problem 48E

The value obtained in part b is different from ln(4) (given ).

Explanation of Solution

Given information:

  $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{2\left(2^h-1\right)}{h}$ and the limit at $h\to 0$ in $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{2\left(2^h-1\right)}{h}$ is 1.39 graphically.

Calculation:

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0}{\lim }\frac{2\left(2^h-1\right)}{h}\\ \text{Apply l'Hopital rule}\\ \underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=2\underset{h\to 0}{\lim }\frac{\frac{d\left(2^h-1\right)}{dh}}{\frac{d\left(h\right)}{dh}}\\ =\underset{h\to 0}{2\lim }\frac{2^h\log 2}{1}\\ \text{Apply limit}\\ \underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=2\log 2\\ =\log 4\end{array}$

Hence the estimated value (using graph) is different from 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 48E

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

Explanation of Solution

Given information:

  $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{2\left(2^h-1\right)}{h}$ .

Concept used:

Find the left hand and right limit of $\frac{f\left(1+h\right)-f\left(1\right)}{h}=\frac{2\left(2^h-1\right)}{h}$ , if the values are equal then slope exist at $x=1$ .

Calculation:

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0^{-}}{\lim }\frac{2\left(2^h-1\right)}{h}\\ \text{Apply l'Hopital rule}\\ \underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=2\underset{h\to 0^{-}}{\lim }\frac{\frac{d\left(2^h-1\right)}{dh}}{\frac{d\left(h\right)}{dh}}\\ =\underset{h\to 0^{-}}{2\lim }\frac{2^h\log 2}{1}\\ \text{Apply limit}\\ \underset{h\to 0^{-}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=2\log 2\\ =\log 4\end{array}$

Now, find left hand limit as:

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{2\left(2^h-1\right)}{h}\\ \text{Apply l'Hopital rule}\\ \underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=2\underset{h\to 0^{+}}{\lim }\frac{\frac{d\left(2^h-1\right)}{dh}}{\frac{d\left(h\right)}{dh}}\\ =\underset{h\to 0^{+}}{2\lim }\frac{2^h\log 2}{1}\\ \text{Apply limit}\\ \underset{h\to 0^{+}}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=2\log 2\\ =\log 4\end{array}$

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

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{h\to 0}{\lim }\frac{2\left(2^h-1\right)}{h}\\ \text{Apply l'Hopital rule}\\ \underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=2\underset{h\to 0}{\lim }\frac{\frac{d\left(2^h-1\right)}{dh}}{\frac{d\left(h\right)}{dh}}\\ =\underset{h\to 0}{2\lim }\frac{2^h\log 2}{1}\\ \text{Apply limit}\\ \underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=2\log 2\\ =\log 4\end{array}$

Hence the slope is ln(4).