Chapter 2.4, Problem 17E

To determine

To check: Whether the curve $f\left(x\right)=\{\begin{array}{l}\frac{1}{x},x\le 2\\ \frac{4-x}{4},x>2\end{array}$ has a tangent at $x=2$ or not. If it exists give the slope of tangent.

Answer to Problem 17E

Yes, the curve $f\left(x\right)=\{\begin{array}{l}\frac{1}{x},x\le 2\\ \frac{4-x}{4},x>2\end{array}$ has a tangent at $x=2$ and the slope of the tangent is $-\frac{1}{4}$.

Explanation of Solution

Given information:

The curve is $f\left(x\right)=\{\begin{array}{l}\frac{1}{x},x\le 2\\ \frac{4-x}{4},x>2\end{array}$ and the point is $x=2$.

Calculation:

For the tangent of the curve $f\left(x\right)$ to exist at $x=a$ , the slope from the left side of the point $x=a$ must be equal to the slope from the right side of the point $x=a$.

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

The slope of the curve from the left side of $x=2$ is:

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

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

  $f\left(2+h\right)=\frac{1}{2+h}$

Substitute $2$ for $x$ in the function $f\left(x\right)=\frac{4-x}{4}$.

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

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

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

Further simplify.

  $\begin{array}{c}\underset{h\to 0^{-}}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=\frac{-1}{2\left(2+0\right)}\\ =\frac{-1}{2\left(2\right)}\\ =-\frac{1}{4}\end{array}$

So, the slope of the curve from the left side of the point $x=2$ is $-\frac{1}{4}$.

The function corresponding to the right side of the point $x=2$ is $f\left(x\right)=\frac{4-x}{4}$.

The slope of the curve from the right side of $x=2$ is:

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

Substitute $2+h$ for $x$ in the function $f\left(x\right)=\frac{4-x}{4}$.

  $\begin{array}{c}f\left(2+h\right)=\frac{4-\left(2+h\right)}{4}\\ f\left(2+h\right)=\frac{4-2-h}{4}\\ f\left(2+h\right)=\frac{2-h}{4}\end{array}$

Substitute $2$ for $x$ in the function $f\left(x\right)=\frac{4-x}{4}$.

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

Substitute $\frac{2-h}{4}$ for $f\left(2+h\right)$ and $\frac{1}{2}$ for $f\left(0\right)$ in the formula for slope.

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{\frac{2-h}{4}-\frac{1}{2}}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{\frac{2-h-2}{4}}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{-h}{4h}\end{array}$

Further simplify.

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{f\left(2+h\right)-f\left(2\right)}{h}=\underset{h\to 0^{+}}{\lim }\left(\frac{-1}{4}\right)\\ =-\frac{1}{4}\end{array}$

So, the slope of the curve from the right side of the point $x=2$ is $-\frac{1}{4}$.

The slope of the curve from the left side of point $x=2$ is equal to the slope from the right side of point $x=2$.

Therefore, the curve $f\left(x\right)=\{\begin{array}{l}\frac{1}{x},x\le 2\\ \frac{4-x}{4},x>2\end{array}$ has a tangent at $x=2$ and the slope of the tangent is $-\frac{1}{4}$.