Chapter 13.3, Problem 14E

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 14E

The equation of the tangent at the point $\left(4,\text{ 3}\right)$ is $y=\frac{x}{3}+\frac{5}{3}$.

Explanation of Solution

Given information:

The given function and the point is $y=\sqrt{1+2x}$ at $\left(4,\text{ 3}\right)$.

Calculation :

The tangent line to the curve $y=f\left(x\right)$ at the point $P\left(a,f\left(a\right)\right)$ is the line through P with slope

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

provided this limit exist.

Let $y=\sqrt{1+2x}$ , then the slope of the tangent line where $a=4$ is

  $\begin{array}{l}m=\underset{h\to 0}{\lim }\frac{f\left(4+h\right)-f\left(4\right)}{h}\\ m=\underset{h\to 0}{\lim }\frac{\sqrt{1+2\left(4+h\right)}-3}{h}\\ m=\underset{h\to 0}{\lim }\frac{\sqrt{2h+9}-3}{h}\cdot \frac{\sqrt{2h+9}+3}{\sqrt{2h+9}+3}\\ m=\underset{h\to 0}{\lim }\frac{2h+9-9}{h\left(\sqrt{2h+9}+3\right)}\\ m=\underset{h\to 0}{\lim }\frac{2\cancel{h}}{\cancel{h}\left(\sqrt{2h+9}+3\right)}\\ m=\underset{h\to 0}{\lim }\frac{2}{\left(\sqrt{2h+9}+3\right)}\\ m=\frac{2}{\left(\sqrt{0+9}+3\right)}\\ m=\frac{1}{3}\end{array}$

Therefore, the equation of the tangent at the point $\left(4,\text{ 3}\right)$ with slope $\frac{1}{3}$ is

  $\begin{array}{l}y-3=\frac{1}{3}\left(x-4\right)\\ y-3=\frac{x}{3}-\frac{4}{3}\\ y=\frac{x}{3}+\frac{5}{3}\end{array}$

The graph of the curve and the tangent line is drawn below:

  

Hence,

The equation of the tangent at the point $\left(4,\text{ 3}\right)$ is $y=\frac{x}{3}+\frac{5}{3}$.