Chapter 1.4, Problem 46E

To determine

To Find: The equation of the tangent to the graph of $f\left(x\right)=\frac{2}{x}$ at $x=1$ .

Answer to Problem 46E

A. $y=\frac{1}{2}x+\frac{3}{2}$ .

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{2}{x}$ and the point is $x=1$ .

Concept used:

The slope of the function $f\left(x\right)$ at any point say $x=a$ , can be calculated as: $\underset{h\to 0}{\lim }{f}^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

If the slope of a line is m, then the slope of the normal will be $-\frac{1}{m}$ . Also, the normal passes through the point $\left(x_0,y_0\right)$ . So, the equation of the normal will be:

  $y-y_0=-\frac{1}{m}\left(x-x_0\right)$ .

Calculation:

The slope of the function $f\left(x\right)=\frac{2}{x}$ at $x=1$ will be:

  $\begin{array}{c}\underset{h\to 0}{\lim }{f}^{\prime }\left(1\right)=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{2}{1+h}-\frac{2}{1}}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{2-2\left(1+h\right)}{1+h}}{h}\\ =\underset{h\to 0}{\lim }\frac{2-2-2h}{h\left(1+h\right)}\\ =\underset{h\to 0}{\lim }\frac{-2h}{h\left(1+h\right)}\\ =\underset{h\to 0}{\lim }\frac{-2}{\left(1+h\right)}\\ =-2\end{array}$

Hence the slope m is $-2$ and $-\frac{1}{m}=\frac{1}{2}$ .

Now find the value of $f\left(x\right)$ at $x=1$ .

  $\begin{array}{c}f\left(1\right)=\frac{2}{1}\\ =2\end{array}$

Now put $\left(x_0,y_0\right)=\left(1,2\right)$ and $m=-2$ in $y-y_0=-\frac{1}{m}\left(x-x_0\right)$ .

  $\begin{array}{c}y-2=\frac{1}{2}\left(x-1\right)\\ y=2+\frac{1}{2}x-\frac{1}{2}\\ y=\frac{1}{2}x+\frac{3}{2}\end{array}$

Hence the equation of the normal line is $y=\frac{1}{2}x+\frac{3}{2}$ .