Question

Find the slope of the tangent line to the graph of fat the given point.$f\left(x\right)=\frac{6}{x+1},at(2,2)$.

Short answer

The slope of the tangent line is $-\frac{1}{2}$.

Step-by-step solution

Step 1. Given information.

The function here given is,

$\begin{array}{l}f\left(x\right)=\frac{6}{x+1}\\ po\mathrm{int}=(2,2)\end{array}$

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$f\left(x\right)=\frac{6}{x+1}$,then the slope of the tangent line at$(2,2)$is,

$\begin{array}{c}m=\underset{x\to 2}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2}\\ =\underset{x\to 2}{\lim }\frac{\left(\frac{6}{x+1}\right)-\left(\frac{6}{2+1}\right)}{x-2}\\ =\underset{x\to 2}{\lim }\frac{\left(\frac{6}{x+1}\right)-\left(\frac{6}{3}\right)}{x-2}\\ =\underset{x\to 2}{\lim }\frac{(\frac{6}{x+1}\cdot \frac{3}{3}-\frac{6}{3}\cdot \frac{x+1}{x+1})}{(x-2)}\\ =\underset{x\to 2}{\lim }\frac{\left(\frac{18-6x-6}{3(x+1)}\right)}{(x-2)}\\ =\underset{x\to 2}{\lim }\frac{\left(\frac{-6x+12}{3(x+1)}\right)}{(x-2)}\\ =\underset{x\to 2}{\lim }\frac{-6\cancel{(x-2)}}{3(x+1)\cancel{(x-2)}}\\ =-\frac{6}{3(2+2)}\\ =-\frac{1}{2}\end{array}$

Therefore, the slope of the tangent line is$-\frac{1}{2}$ .