Chapter 2.4, Problem 16E

To determine

To check: Whether the curve $f\left(x\right)=\{\begin{array}{l}-x,x<0\\ x^2-x,x\ge 0\end{array}$ has a tangent at $x=0$ or not. If it exists give the slope of tangent.

Answer to Problem 16E

Yes, the curve $f\left(x\right)=\{\begin{array}{l}-x,x<0\\ x^2-x,x\ge 0\end{array}$ has a tangent at $x=0$ and the slope of the tangent is $-1$.

Explanation of Solution

Given information:

The curve is $f\left(x\right)=\{\begin{array}{l}-x,x<0\\ x^2-x,x\ge 0\end{array}$ and the point is $x=0$.

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 $x=0$ is $f\left(x\right)=-x$.

The slope of the curve from the left side of $x=0$ is:

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

Substitute $h$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=-x\\ f\left(h\right)=-h\end{array}$

Substitute $0$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=x^2-x\\ f\left(0\right)={\left(0\right)}^2-0\\ f\left(0\right)=0\end{array}$

Substitute $-h$ for $f\left(h\right)$ and $0$ for $f\left(0\right)$ in the formula for slope.

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

So, the slope of the curve from the left side of the point $x=0$ is $-1$.

The function corresponding to the right side of the point $x=0$ is $f\left(x\right)=x^2-x$.

The slope of the curve from the right side of $x=0$ is:

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

Substitute $h$ for $x$ in the function.

  $\begin{array}{l}f\left(x\right)=x^2-x\\ f\left(h\right)=h^2-h\end{array}$

Substitute $0$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=x^2-x\\ f\left(0\right)={\left(0\right)}^2-0\\ f\left(0\right)=0\end{array}$

Substitute $h^2-h$ for $f\left(h\right)$ and $0$ for $f\left(0\right)$ in the formula for slope.

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

So, the slope of the curve from the right side of the point $x=0$ is $-1$.

The slope of the curve $f\left(x\right)=\{\begin{array}{l}-x,x<0\\ x^2-x,x\ge 0\end{array}$ from the left side of point $x=0$ is equal to the slope from the right side of point $x=0$.

Therefore, the curve $f\left(x\right)=\{\begin{array}{l}-x,x<0\\ x^2-x,x\ge 0\end{array}$ has a tangent at $x=0$ and the slope of the tangent is $-1$.