How do you find the intervals of increasing and decreasing using the first derivative given #y=abs(x+4)-1#?

2 Answers
Jan 4, 2018

Increasing when #x>=-4# and decreasing when #x<-4#

Explanation:

Let's first rewrite this function as a piece wise one.
We have:
#y=x+4-1=>x+3# when #x>=-4#
#y=-(x+4)-1=>y=-x-5# when #x<-4#

Now, we try to find the derivatives for each case. We therefore have:
#x+3=>1#
#-x-5=>-1#
This really tells us that whenever #x# is greater than or equal to #-4#, the rate is always increasing.
Similarly, whenever #x# is less than #-4#, the rate is always decreasing.

We can see that when we graph this:
graph{abs(x+4)-1 [-10, 10, -5, 5]}

We can see that there is a "break" at #x=-4#.
And we can see that the line goes downward to the left of #-4# and upward to the right of #-4#.

Jan 4, 2018

#f# is strictly increasing in #[-4,+oo)# , strictly decreasing in #(-oo,-4]#

Explanation:

#f(x)=|x+4|-1#,
#D_f=RR#

  • If #x+4>0# #<=># #x>##-4# then

#f(x)=x+4-1=x+3#

and #f'(x)=1# ,

  • If #x+4<0# #<=># #x<##-4# then

#f(x)=-x-4-1=-x-5#

and #f'(x)=-1#,

Therefore,

#f(x) = {(x+3", "x> -4),(-x-5" , "x<=-4):}#

&

#f'(x) = {(1", "x> -4),(-1" , "x<=-4):}#

so #f'(x)>0# , #x##in##(-4,+oo)# so #f# is strictly increasing in #(-4,+oo)#

#f'(x)<0# , #x##in##(-oo,-4)# so #f# is strictly decreasing in #(-oo,-4]#