Chapter 3.2, Problem 15E

To determine

Whether the given function $y$ is a corner, a cusp, a vertical tangent, or a discontinuity.

Answer to Problem 15E

The given function $y$ is a corner.

Explanation of Solution

Given information:

The given function is $y=3x-2\left|x\right|-1$ and which is fails to be differentiable at $x=0$ .

Formula Used:

Differentiable function formula:

The differentiable function formulas to find the left hand derivative and right hand derivative are

Left hand derivative $=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ and Right hand derivative $=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ respectively.

Calculation:

The given function is:

  $y=3x-2\left|x\right|-1$

The given function can be written as:

  $\begin{array}{c}y=\{\begin{array}{l}3x-2x-1,x\ge 0\\ 3x-2\left(-x\right)-1,x<0\end{array}\\ =\{\begin{array}{c}x-1,x\ge 0\\ 5x-1,x<0\end{array}\end{array}$

To find the Left hand derivative, use the differentiable function formula. $\text{Left}\text{hand}\text{derivative}=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ .

Substitute $y=x-1$ for $f\left(x\right)$ and $0$ for $a$ in the above formula.

  $\begin{array}{c}\text{Left}\text{hand}\text{derivative}=\underset{x\to 0^{-}}{\lim }\frac{\left(x-1\right)-f\left(0\right)}{x-0}\\ =\underset{x\to 0^{-}}{\lim }\frac{\left(x-1\right)-\left(-1\right)}{x}\\ =\underset{x\to 0^{-}}{\lim }\frac{x-1+1}{x}\\ =\underset{x\to 0^{-}}{\lim }\frac{x}{x}\end{array}$

Simplify the above limit further.

  $\begin{array}{c}\underset{x\to 0^{-}}{\lim }\frac{x}{x}=\underset{x\to 0^{-}}{\lim }1\\ =1\end{array}$

Again, to find the Right hand derivative, use the differentiable formula $\text{Right}\text{hand}\text{derivative}=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ .

Substitute $5x-1$ for $f\left(x\right)$ and $0$ for $a$ in the above formula.

  $\begin{array}{c}\text{Right}\text{hand}\text{derivative}=\underset{x\to 0^{+}}{\lim }\frac{\left(5x-1\right)-f\left(0\right)}{x-0}\\ =\underset{x\to 0^{+}}{\lim }\frac{5x-1-\left(-1\right)}{x}\\ =\underset{x\to 0^{-}}{\lim }\frac{5x-1+1}{x}\\ =\underset{x\to 0^{+}}{\lim }\frac{5x}{x}\end{array}$

Simplify the above limit further.

  $\begin{array}{c}\underset{x\to 0^{+}}{\lim }\frac{5x}{x}=\underset{x\to 0^{+}}{\lim }5\\ =5\end{array}$

It is seen that the slopes on the left hand derivative and right hand derivative at $x=0$ are $1$ and $5$ respectively which is enough for the function not to be differentiable at $x=0$ .

Thus, the results of the limit indicate that the given function is a corner.

The graph of the function $y=3x-2\left|x\right|-1$ has a vertical at $x=0$

  

Hence, the given function $y=3x-2\left|x\right|-1$ is a corner.