   Chapter 3.1, Problem 74E

Chapter
Section
Textbook Problem

Where is the function h(x) = |x – 1| + |x + 2| differentiable? Give a formula for h' and sketch the graphs of hand h'.

To determine

To find: The value of x where the function h(x) is differentiable and obtain the formula for h(x); sketch the graph of the function h(x) and h(x).

Explanation

Given:

The function h(x)=|x1|+|x+2|.

Derivative rules:

(1) Power Rule: ddx(xn)=nxn1

(2) Constant multiple Rule: ddx[cf]=cddx(f)

(3) Sum Rule: ddx[f(x)+g(x)]=ddx(f(x))+ddx(g(x))

Calculation:

Express the given function and express h(x) as follows.

h(x)={(x1)(x+2)if x2(x1)+(x+2)if 2<x<1(x1)+(x+2)if x1={2x1if  x23if 2<x<12x+1 if  x1

Obtain the left hand derivative of h(x) at 2.

That is, compute limk0h(2+k)h(2)k.

Since h(x)=2x1 when x2,

limk0h(2+k)h(2)k=limk0h(2+k)h(2)k=limk0[2(2+k)1][2(2)1]k=limk0[42k1]k

Perform the mathematical operation and simplify the terms,

limk0h(2+k)h(2)k=limk032k3k=limk02kk=limk0(2)=2

Thus, the value of the left hand derivative of h(x) at 2 is 2.

Obtain the right hand limit of h(x) at 2.

That is, compute limk0+h(2+k)h(2)k.

Since h(x)=3 when x>2,

limk0+h(2+k)h(2)k=limk0+h(2+k)h(2)k=limk0+(33)k=limk0+(0)=0

Thus, the value of the right hand derivative of h(x) at 2 is 0.

Since the value of the left hand derivative of h(x) and the value of the right hand derivative of h(x) are not equal, the limit h(2)=limk0h(2+k)h(2)k does not exist

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