Chapter 2.2, Problem 15E

To determine

To find: Weather the given function has a corner, cusp, vertical tangent and discontinuity.

Answer to Problem 15E

The function has a corner.

The function has corner.

Explanation of Solution

Given information:

The function is $y=3x-2\left|x\right|-1$ , the function is not differentiable at $x=0$ .

Concept used:

If a function is continuous then it can have cusp, corner or vertical tangent otherwise it will be discontinuous.

Calculation:

The function is $y=3x-2\left|x\right|-1$ .

The above function can be written in the form of piecewise function as shown below.

  $\begin{array}{c}y=\left\{\begin{array}{l}3x-2x-1,x\le 0\\ 3x+2x-1,x>0\end{array}\right\}\\ =\left\{\begin{array}{l}x-1,x\le 0\\ 5x-1,x>0\end{array}\right\}\end{array}$

Find the right hand derivative.

  $\begin{array}{c}\underset{h\to 0^{+}}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=\underset{h\to 0^{+}}{\lim }\frac{\left[5\left(0+h\right)-1\right]-\left[5\left(0\right)-1\right]}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{5h-1+1}{h}\\ =\underset{h\to 0^{+}}{\lim }\frac{5h}{h}\\ =5\end{array}$

Find the left hand derivative.

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

Since the left hand derivative and right hand derivative are definite but not equal.

Conclusion: So, the function has a corner.