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

Question

William Abdalla · Accepted Answer

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

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.

When we graph this, we can observe that: 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#.

Emma Aldridge · Answer

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

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

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

and #f'(x)=1# ,

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

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

Consequently,

#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]#

Emily Ammerman · Answer

undefined To find the intervals of increasing and decreasing for the function $ y = |x + 4| - 1 $ using its first derivative:

1. Find the first derivative of the function $ y = |x + 4| - 1 $.
   $ y' = \frac{{d}}{{dx}}(|x + 4| - 1) $

2. Split the derivative into two cases: 
   - When $ x + 4 \geq 0 $, the derivative is $ y' = 1 $.
   - When $ x + 4 < 0 $, the derivative is $ y' = -1 $.

3. Set each derivative equal to zero to find critical points:
   - For $ y' = 1 $, there are no critical points since it's a constant.
   - For $ y' = -1 $, there are no critical points since it's a constant.

4. Determine the intervals where the derivative is positive or negative:
   - $ y' = 1 $ is positive for all $ x \geq -4 $.
   - $ y' = -1 $ is negative for all $ x < -4 $.

5. Therefore, the function is increasing for $ x \geq -4 $ and decreasing for $ x < -4 $.