Chapter 12, Problem 62RE

a.

To determine

Find a Formula for the slope of a graph.

Answer to Problem 62RE

Slope at points $\left(1,1\right)$ is $\boxed{\frac{1}{2}}$

Explanation of Solution

Given information:

Find a formula for the slope of the graph of $f$ at the points $\left(x,f\left(x\right)\right)$ .Then use it to find the slope at the two given points.

  $f\left(x\right)=\sqrt{x}$ , point is $\left(1,1\right)$

Calculation:

Consider the function,

  $f\left(x\right)=x^2-4x$

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

  $m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Determine the limit $m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

  $\begin{array}{l}m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}\end{array}$

Apply rationalization.

  $\begin{array}{l}=\underset{h\to 0}{\lim }\frac{\left(\sqrt{x+h}-\sqrt{x}\right)\left(\sqrt{x+h}+\sqrt{x}\right)}{\left(\sqrt{x+h}+\sqrt{x}\right)h}\\ =\underset{h\to 0}{\lim }\frac{{\left(\sqrt{x+h}\right)}^2-\left(\sqrt{x}\right)}{\left(\sqrt{x+h}+\sqrt{x}\right)h}\\ =\underset{h\to 0}{\lim }\frac{h}{\left(\sqrt{x+h}+\sqrt{x}\right)h}\end{array}$

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

  $\begin{array}{l}=\underset{h\to 0}{\lim }\frac{\cancel{h}}{\left(\sqrt{x+h}+\sqrt{x}\right)\cancel{h}}=\underset{h\to 0}{\lim }\frac{1}{\left(\sqrt{x+h}+\sqrt{x}\right)}\\ =\underset{h\to 0}{\lim }\frac{1}{\left(\sqrt{x+0}+\sqrt{x}\right)}\\ =\frac{1}{2\sqrt{x}}\end{array}$

Therefore slope of $f$ at the points $\left(x,f\left(x\right)\right)$ is $\frac{1}{2\sqrt{x}}$

Now calculate the slope at points $\left(1,1\right)$

The slope is

  $m=\frac{1}{2\sqrt{x}}$

Substitute the value of $x=1$ in the slope,

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

Hence slope at points $\left(1,1\right)$ is $\boxed{\frac{1}{2}}$

b.

To determine

Find a Formula for the slope of a graph.

Answer to Problem 62RE

Slope at points $\left(4,2\right)$ is $\boxed{\frac{1}{4}}$

Explanation of Solution

Given information:

Find a formula for the slope of the graph of $f$ at the points $\left(x,f\left(x\right)\right)$ .Then use it to find the slope at the two given points.

  $f\left(x\right)=\sqrt{x}$ , point is $\left(4,2\right)$

Calculation:

Consider the function,

  $f\left(x\right)=x^2-4x$

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

  $m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

Determine the limit $m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

  $\begin{array}{l}m=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}\end{array}$

Apply rationalization.

  $\begin{array}{l}=\underset{h\to 0}{\lim }\frac{\left(\sqrt{x+h}-\sqrt{x}\right)\left(\sqrt{x+h}+\sqrt{x}\right)}{\left(\sqrt{x+h}+\sqrt{x}\right)h}\\ =\underset{h\to 0}{\lim }\frac{{\left(\sqrt{x+h}\right)}^2-\left(\sqrt{x}\right)}{\left(\sqrt{x+h}+\sqrt{x}\right)h}\\ =\underset{h\to 0}{\lim }\frac{h}{\left(\sqrt{x+h}+\sqrt{x}\right)h}\end{array}$

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

  $\begin{array}{l}=\underset{h\to 0}{\lim }\frac{\cancel{h}}{\left(\sqrt{x+h}+\sqrt{x}\right)\cancel{h}}=\underset{h\to 0}{\lim }\frac{1}{\left(\sqrt{x+h}+\sqrt{x}\right)}\\ =\underset{h\to 0}{\lim }\frac{1}{\left(\sqrt{x+0}+\sqrt{x}\right)}\\ =\frac{1}{2\sqrt{x}}\end{array}$

Therefore slope of $f$ at the points $\left(x,f\left(x\right)\right)$ is $\frac{1}{2\sqrt{x}}$

Now calculate the slope at points $\left(4,2\right)$

The slope is

  $m=\frac{1}{2\sqrt{x}}$

Substitute the value of $x=4$ in the slope,

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

Hence slope at points $\left(4,2\right)$ is $\boxed{\frac{1}{4}}$