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.
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