Chapter 11, Problem 62RE

(a)

To determine

To calculate:

The formula for the slope of graph and then use it to get the slope of specified points.

Answer to Problem 62RE

The formula for the slope of graph is $m=\frac{1}{2\sqrt{x}}$ and then use it to get the slope of specified points is $\frac{1}{2}$ .

Explanation of Solution

Given information:

Point: $\left(x,f\left(x\right)\right)$ , $\left(1,1\right)$

Function: $f\left(x\right)=\sqrt{x}$

Calculation:

Consider the following function

  $f\left(x\right)=\sqrt{x}$

the slope of the graph of $f$ at the $\left(x,f\left(x\right)\right)$ is

  $m=\underset{x\to \infty }{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ , provided the limit exists.

Get the limit $m=\underset{x\to \infty }{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

  $\underset{x\to \infty }{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}=\underset{x\to \infty }{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}$

Now, multiply both the numerator and the denominator by $\left(\sqrt{x+h}-\sqrt{x}\right)$

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}=\underset{x\to \infty }{\lim }\frac{\left(\sqrt{x+h}-\sqrt{x}\right)\left(\sqrt{x+h}+\sqrt{x}\right)}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\\ =\underset{x\to \infty }{\lim }\frac{{\left(\sqrt{x+h}\right)}^2-{\left(\sqrt{x}\right)}^2}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\\ =\underset{x\to \infty }{\lim }\frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\end{array}$

Since, $h\to 0$ , so $h\ne 0$

Divide both the numerator and the denominator by $h$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{1}{2\sqrt{x}}$

To get a slope of the graph of $f$ at the point $\left(1,1\right)$ , substitute $x=1$ in $m=\frac{1}{2\sqrt{x}}$

Therefore, the slope of $f$ at the point $\left(1,1\right)$ is

  $\begin{array}{l}m=\frac{1}{2\sqrt{1}}\\ =\frac{1}{2}\end{array}$

(b)

To determine

To calculate:

The formula for the slope of graph and then use it to get the slope of specified points.

Answer to Problem 62RE

The formula for the slope of graph is $m=\frac{1}{2\sqrt{x}}$ and then use it to get the slope of specified points is $\frac{1}{4}$ .

Explanation of Solution

Given information:

Point: $\left(x,f\left(x\right)\right)$ , $\left(4,2\right)$

Function: $f\left(x\right)=\sqrt{x}$

Calculation:

Consider the following function

  $f\left(x\right)=\sqrt{x}$

the slope of the graph of $f$ at the $\left(x,f\left(x\right)\right)$ is

  $m=\underset{x\to \infty }{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$ , provided the limit exists.

Get the limit $m=\underset{x\to \infty }{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

  $\underset{x\to \infty }{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}=\underset{x\to \infty }{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}$

Now, multiply both the numerator and the denominator by $\left(\sqrt{x+h}-\sqrt{x}\right)$

  $\begin{array}{l}\underset{x\to \infty }{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}=\underset{x\to \infty }{\lim }\frac{\left(\sqrt{x+h}-\sqrt{x}\right)\left(\sqrt{x+h}+\sqrt{x}\right)}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\\ =\underset{x\to \infty }{\lim }\frac{{\left(\sqrt{x+h}\right)}^2-{\left(\sqrt{x}\right)}^2}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\\ =\underset{x\to \infty }{\lim }\frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}\end{array}$

Since, $h\to 0$ , so $h\ne 0$

Divide both the numerator and the denominator by $h$ at the point $\left(x,f\left(x\right)\right)$ is $m=\frac{1}{2\sqrt{x}}$

To get a slope of the graph of $f$ at the point $\left(4,2\right)$ , substitute $x=4$ in $m=\frac{1}{2\sqrt{x}}$

Therefore, the slope of $f$ at the point $\left(4,2\right)$ is

  $\begin{array}{l}m=\frac{1}{2\sqrt{4}}\\ =\frac{1}{4}\end{array}$