Question

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

Answer 1

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`.

Answer 2

`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]`