Chapter 13, Problem 34RE

To determine

To find an equation of the line tangent to the graph of $f$ at the given point

Answer to Problem 34RE

The equation is $x-4y=-5$

Explanation of Solution

Given information:

Consider the given information,

  $f\left(x\right)\sqrt{x+1}\text{ at }\left(3,2\right)$

CALCULATION:

To find the equation is,

  $\begin{array}{l}f\left(x\right)=\sqrt{x+1}\\ f\text{'}\left(x\right)=\frac{1}{2}{\left(x+1\right)}^{\frac{1}{2}-1}\\ \text{At }\left(3,2\right)f\text{'}\left(3\right)=\frac{1}{2\sqrt{3+1}}=\frac{1}{2\sqrt{4}}=\frac{1}{4}\\ \text{Theefore,}m=\frac{1}{4}\\ \text{equation,}\left(y-y_1\right)=m\left(x-x_1\right)\\ =\left(y-2\right)=\frac{1}{4}\left(x-3\right)\\ =4y-8=x-3\\ =x-4y=-5\end{array}$

Therefore, the equation of the line tangent to the graph of $f$ at the given point is $x-4y=-5$