This question aims to find the interval of **increase** or interval of **decrease** of the given function by finding its **critical points** first.

The interval of increase and decrease is the interval in which the real function will increase or decrease in the value of a **dependent variable.** The increase or decrease of the interval can be found by checking the value of the **first derivative** of the given function.

If the derivative is **positive**, this means that the interval is increasing. It implies the increase of function with the dependent variable $ x $. If the derivative is **negative**, this means that the interval is decreasing. It implies the decrease of function with the dependent variable x .

## Expert Answer

Let the function be:

\[f(x) = x ^\frac{1}{5} ( x + 6 ) \]

Taking **first derivative** of the function $f (x)$:

\[f’ (x) =\frac{1}{5} \pi ^ \frac{-4}{5} ( x + 6 ) + x^ \frac{1}{5}\]

\[=\frac{x + 6}{5x ^ {\frac{4}{5}}} + x ^\frac{1}{5}\]

\[=\frac{ x + 6 + 5x ^ {\frac{1}{5}+ \frac{4}{5}}}{ 5x^{\frac{4}{5}} }\]

Taking $6$ common, we get:

\[=\frac{6 (x + 1) }{ 5x ^ {\frac{4}{5}}}\]

To find critical points, we will put the first derivative equal to $0$:

\[f’ (x) = 0\]

\[\frac{ 6 (x + 1) }{ 5x ^ {\frac{4}{5}} } = 0\]

\[x + 1 = 0\]

\[x = – 1\]

**The critical points are $x = – 1$ and $x = 0$**

The interval is then:

\[(- \infty , – 1 ) , (- 1 , 0) , (0 , \infty)\]

## Numerical Solution

In the given interval $( – \infty , – 1 )$, put $x = -2$

\[\frac{ 6 (- 2 + 1) }{ 5( – 2) ^ {\frac{4}{5}} } = – 0 . 68 < 0\]

**Thus, $f (x)$ is decreasing in the interval $(- \infty , – 1)$.**

Take the interval $( -1 , 0 )$ and put $x = – 0.5$:

\[f’ (x) = \frac{ 6 ( – 0.5 + 1) }{ 5( – 0.5 ) ^ {\frac{4}{5}} } = 1.04 > 0\]

**So $f (x)$ is increasing in the interval $( – 1 , 0 )$.**

In the interval $(0 , \infty)$, put $x = 1$:

\[f’ (x) =\frac{6 ( 1 + 1) }{5( 1) ^ {\frac{4}{5}}} = 2.4 > 0\]

**So $f(x)$ is increasing in the interval $(0 , \infty)$.**

## Example

**Find the increasing and decreasing intervals of the function $f(x)= -x^3 + 3x^2 +9$.**

\[f’(x) = -3x^2 + 6x\]

\[f’(x) = -3x (x – 2)\]

To find critical points:

\[-3x (x – 2) = 0\]

$x = 0$ or $x = 2$

The intervals are $(- \infty, 0)$ , $(0, 2)$ and $(2, \infty)$.

For interval $(- \infty , 0 )$, put $x = -1$:

\[f’ (x) = -9 < 0\]

**It is a decreasing function**.

For interval $(0, 2)$, put $x =1$:

\[f’ (x) = 3 > 0\]

**It is an increasing function.**

For interval $(2, \infty)$, put $x =4$:

\[f’ (x) = -24 < 0\]

**It is a decreasing function**.

*Image/Mathematical drawings are created in Geogebra.*