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

Jan 4, 2018

Increasing when $x \ge - 4$ and decreasing when $x < - 4$

#### Explanation:

Let's first rewrite this function as a piece wise one.
We have:
$y = x + 4 - 1 \implies x + 3$ when $x \ge - 4$
$y = - \left(x + 4\right) - 1 \implies y = - x - 5$ when $x < - 4$

Now, we try to find the derivatives for each case. We therefore have:
$x + 3 \implies 1$
$- x - 5 \implies - 1$
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 $\left[- 4 , + \infty\right)$ , strictly decreasing in $\left(- \infty , - 4\right]$

#### Explanation:

$f \left(x\right) = | x + 4 | - 1$,
${D}_{f} = \mathbb{R}$

• If $x + 4 > 0$ $\iff$ $x >$$- 4$ then

$f \left(x\right) = x + 4 - 1 = x + 3$

and $f ' \left(x\right) = 1$ ,

• If $x + 4 < 0$ $\iff$ $x <$$- 4$ then

$f \left(x\right) = - x - 4 - 1 = - x - 5$

and $f ' \left(x\right) = - 1$,

Therefore,

$f \left(x\right) = \left\{\begin{matrix}x + 3 \text{ & "x> -4 \\ -x-5" & } x \le - 4\end{matrix}\right.$

&

$f ' \left(x\right) = \left\{\begin{matrix}1 \text{ & "x> -4 \\ -1" & } x \le - 4\end{matrix}\right.$

so $f ' \left(x\right) > 0$ , $x$$\in$$\left(- 4 , + \infty\right)$ so $f$ is strictly increasing in $\left(- 4 , + \infty\right)$

$f ' \left(x\right) < 0$ , $x$$\in$$\left(- \infty , - 4\right)$ so $f$ is strictly decreasing in $\left(- \infty , - 4\right]$