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M 408C
Spring 2002
Homework #8, Part II
In each of the problems below, differentiate the given function f(x). Since the given function f(x) is defined
as the absolute value of another function, you will first need to express the formula for f(x) with a piece-wise
defined formula which does not involve the absolute value function. Then, differentiate the functions which
define f(x) on the pieces of the domain individually. The formula for the derivative f’(x) will then be a piecewise defined formula, defined by the resulting derivatives on pieces of the domain. The pieces of the domain of
f’(x) will differ (if at all) from the pieces of the domain of f(x) only in the inclusion or exclusion of the
endpoints of those pieces.
The value of f’(x) at the endpoints of the pieces of the domain will depend on whether the formulas for f’(x)
on both sides of the point evaluate to the same one-sided derivative at that point (in which case, this common
one-sided derivative value is also the value of f’(x) at that endpoint) or whether these formulas for f’(x)
evaluate to two different one-sided derivative values (in which case, f is not differentiable at that endpoint, f’(x)
is undefined at that endpoint, and that endpoint should not be included in the domain of definition of f’(x) ).
Example Problem and Solution:
Let f ( x) 
x 3  16 x . Differentiate the function f .
Solution: Recalling the steps presented in the assignment for HW #6, Part II:
Step 1) u  x 3  16 x  x x  4 x  4 .
Step 2) u = 0: x ( x + 4 ) ( x – 4 ) = 0 ; x = 0 , – 4 , 4
Step 3) Intervals: ( –infinity, –4 ) ; ( –4, 0 ) ; ( 0, 4 ) ; ( 4, infinity )
Steps 4 and 5:
Test ( –infinty, –4 ) : x < –4 .
At x = –5 , u = –45 < 0 ; |u| = –u .
Test ( –4, 0 ) : –4 < x < 0 .
At x = –2 , u = 24 > 0 ; |u| = +u .
Test ( 0, 4 ) : 0 < x < 4 .
At x = 2 , u = –24 < 0 ; |u| = –u .
Test ( 4, +infinity ) : x > 4 .
At x = 5 , u = 45 > 0 ; |u| = +u .
Step 6): Since u = 0 at the three endpoints, x = –4 , x = 0, and x = 4, extend the use of the |u| = +u formula
from ( –4, 0 ) to the closed interval [ –4 , 0] , and extend the use of the |u| = +u formula from ( 4, infinity ) to
the left-closed interval [ –4 , infinity).
The piece-wise defined formula for f(x) is:
  x 3  16 x 


  x 3  16 x 

f ( x)  x 3  16 x
 
  x 3  16 x 



  x 3  16 x 



3
 x  16 x for  4  x  0 


3
 16 x  x
for 0  x  4 



 x 3  16 x for x  4
 16 x  x 3 for x   4
Differentiating f over each piece of the domain individually, we see that
f ( x)  3x 2  16 or
f ( x)  16  3x 2 depending on which piece of the domain contains the number x .
In considering the definition of f’(x) at the endpoints x = –4 , x = 0 , and x = 4 , we conclude that,
because 16 – 3(–4)2 = –32 and 3(–4)2 – 16 = 32 and these one-sided derivative values differ on either side
of x = –4, f is not differentiable at x = –4 , and so x = –4 should not be included in the domain of
definition of f’(x). Similarly, the one-sided derivative values differ on either side of x = 0 (those values being
16 and –16) and so x = 0 should not be included in the domain of definition of f’(x). Also similarly, x = 4
should not be included in the domain of definition of f’(x). (Check for yourself that the one-sided derivative
values on either side of x = 4 are –32 and 32 .)
Thus, the piece-wise defined formula for f’(x) is as follows:
f ( x) 
d
dx

x 3  16 x













16  3 x 2
3 x 2  16
16  3 x 2
3 x 2  16



for  4  x  0 


for 0  x  4 



for x  4
for x   4
HW #8, Part II Problems:
In each problem, differentiate the given function f(x). The formula for f’(x) will be a piece-wise defined
formula, defined on pieces of the domain which are similar to the pieces of the domain used in the piece-wise
defined formula for the given function f(x). The pieces of the domain of definition of f’(x) will differ (if at all)
from the pieces of the domain of definition of f(x) only in the inclusion or exclusion of the endpoints of these
intervals.
Problems 1 and 2:
Problem 1:
f ( x) 
x 2  10 x  16
Problem 2:
f ( x) 
1
x