Question

Find the slope of the tangent line to the graph of rat the given point.$f\left(x\right)=\frac{5}{x+2},at(3,1)$.

Short answer

The slope of the tangent line is$-\frac{1}{5}$ .

Step-by-step solution

Step 1. Given information.

The function here given is,

$\begin{array}{l}f\left(x\right)=\frac{5}{x+2}\\ po\mathrm{int}=(3,1)\end{array}$

Step 2. Formula used.

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

$m=\underset{x\to a}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$.

provided that this limit exists.

Step 3. Finding the slope of tangent line at given point.

Let$f\left(x\right)=\frac{5}{x+2}$,then the slope of the tangent line at$(3,1)$is,

$\begin{array}{c}m=\underset{x\to 3}{\lim }\frac{f\left(x\right)-f\left(3\right)}{x-3}\\ =\underset{x\to 3}{\lim }\frac{\left(\frac{5}{x+2}\right)-\left(\frac{5}{3+2}\right)}{x-1}\\ =\underset{x\to 3}{\lim }\frac{\left(\frac{5}{x+2}\right)-\left(\frac{5}{5}\right)}{x-1}\\ =\underset{x\to 3}{\lim }\frac{(\frac{5}{x+2}\cdot \frac{3}{3}-\frac{5}{5}\cdot \frac{x+2}{x+2})}{(x-1)}\\ =\underset{x\to 3}{\lim }\frac{\left(\frac{15-5x-10}{5(x+2)}\right)}{(x-1)}\\ =\underset{x\to 3}{\lim }\frac{\left(\frac{-5x+5}{5(x+2)}\right)}{(x-1)}\\ =\underset{x\to 3}{\lim }\frac{-5\cancel{(x-1)}}{5(x+2)\cancel{(x-1)}}\\ =-\frac{5}{5(3+2)}\\ =-\frac{1}{5}\end{array}$

Therefore, the slope of the tangent line is $-\frac{1}{5}$.