Chapter 2.4, Problem 43E

To determine

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

Answer to Problem 43E

The correct choice is (C) such that the equation of the tangent to the graph of $f\left(x\right)=\frac{2}{x}$ at $x=1$ is $y=-2x+4$.

Explanation of Solution

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

The formula for the slope of the curve $f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ is equal to $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$.

The slope of $f\left(x\right)=\frac{2}{x}$ at $x=1$ is:

  $\text{slope}=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}$

Substitute $1+h$ for $x$ in the function $f\left(x\right)=\frac{2}{x}$.

  $f\left(1+h\right)=\frac{2}{1+h}$

Substitute $1$ for $x$ in the function $f\left(x\right)=\frac{2}{x}$.

  $f\left(1\right)=2$

Substitute $\frac{2}{1+h}$ for $f\left(1+h\right)$ and $2$ for $f\left(1\right)$ in the formula for slope.

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

Further simplify.

  $\begin{array}{c}\text{slope}=\frac{-2}{1+0}\\ =\frac{-2}{1}\\ =-2\end{array}$

So, the slope of the curve $f\left(x\right)=\frac{2}{x}$ at $x=1$ is $-2$.

The equation of the tangent line to $y=f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ with slope at $x=a$ is:

  $y-f\left(a\right)=m\left(x-a\right)$

Substitute $-2$ for $m$ , $1$ for $a$ and $2$ for $f\left(1\right)$ in the equation of tangent.

  $\begin{array}{c}y-f\left(1\right)=-2\left(x-1\right)\\ y-2=-2x+2\\ y=-2x+2+2\\ y=-2x+4\end{array}$

Therefore, the correct choice is (C) such that the equation of the tangent to the graph of $f\left(x\right)=\frac{2}{x}$ at $x=1$ is $y=-2x+4$.