Chapter 13.3, Problem 20E

To determine

To find : An equation of the tangent line to the curve at the given point and then graph the curve and the tangent line.

Answer to Problem 20E

The derivative of the function at 4 is $\frac{1}{2}$.

Explanation of Solution

Given information:

The given function and the point is $G\left(x\right)=1+2\sqrt{x}$ at 4.

Calculation :

The derivative of a function $f$ at a number $a$ , denoted by $f\text{'}\left(a\right)$ ,is

  $f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$

provided this limit exist.

According to definition, the derivative of $G\left(x\right)=1+2\sqrt{x}$ , with $a=4$ is

  $\begin{array}{l}G\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{f\left(4+h\right)-f\left(4\right)}{h}\\ G\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{1+2\sqrt{4+h}-\left(1+2\sqrt{4}\right)}{h}\\ G\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{1+2\sqrt{4+h}-5}{h}\\ G\text{'}\left(1\right)=\underset{h\to 0}{\lim }\frac{2\sqrt{4+h}-4}{h}\\ G\text{'}\left(1\right)=2\underset{h\to 0}{\lim }\frac{\sqrt{4+h}-2}{h}\cdot \frac{\sqrt{4+h}+2}{\sqrt{4+h}+2}\\ G\text{'}\left(1\right)=2\underset{h\to 0}{\lim }\frac{4+h-4}{h\left(\sqrt{4+h}+2\right)}\\ G\text{'}\left(1\right)=2\underset{h\to 0}{\lim }\frac{\cancel{h}}{\cancel{h}\left(\sqrt{4+h}+2\right)}\\ G\text{'}\left(1\right)=2\times \frac{1}{\left(\sqrt{4+0}+2\right)}\\ G\text{'}\left(1\right)=2\times \frac{1}{4}\\ G\text{'}\left(1\right)=\frac{1}{2}\end{array}$

Hence,

The derivative of the function at 4 is $\frac{1}{2}$.