Chapter 1.4, Problem 16E

To determine

To Explain: If the curve has tangent at the given point.

Answer to Problem 16E

Yes, the tangent at the given point exists and the slope is $-1$ .

Explanation of Solution

Given information:

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

Concept used:

If the curve has a tangent at any point say $x=a$, then the two-sided limit of the difference quotient should exist, that is:

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

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

Calculation:

First check if the two-sided quotient limit of $f\left(x\right)=\left\{\begin{array}{l}-x,x<0\\ x^2-x,x\ge 0\end{array}\right.$ at $x=0$ exist or not.

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

Here it can be seen that $\underset{h\to 0^{-}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ . Hence the tangent at $x=0$ exist.

The slope at $x=0$ will be $\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ .

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

The slope is $-1$ .