Question

Find the intervals on which $f\left(x\right)=x^4-8x^2$ is increasing and decreasing.

Answer 1

Step 1. Find the critical point

Given function: $f\left(x\right)=x^4-8x^2$

The first derivative of the given function will be,

$$\begin{gathered} f\text{'}\left(x\right)=4x^3-8\left(2\right)x \\ \Rightarrow f\text{'}\left(x\right)=4x^3-16x \end{gathered}$$

Now, we equate $f\text{'}\left(x\right)$to zero.

$$\begin{gathered} 4x^3-16x=0 \\ \Rightarrow 4x(x^2-4)=0 \\ \Rightarrow 4x(x-2)(x+2)=0\mathbf{}[\because a^2-b^2=(a-b\left)\right(a+b\left)\right] \end{gathered}$$

Hence,the critical points are $x=0,x=-2$and $x=2$.

Step 2. Find the interval of increase and decrease

So, the four intervals formed here will be, $(-\infty ,-2),(-2,0),(0,2)$ and $(2,\infty )$

Now, we can identify the increasing and decreasing range by putting values of these intervals in $f\text{'}\left(x\right)$.

Replace, $-1\in \left(-2,0\right)$

$$\begin{gathered} f\text{'}(-1)=4{\left(-1\right)}^3-16\left(-1\right) \\ =-4+16 \\ =8>0 \end{gathered}$$

and, $3\in \left(2,\infty \right)$

$$\begin{gathered} f\text{'}\left(3\right)=4{\left(3\right)}^3-16\left(3\right) \\ =108-48 \\ =60>0 \end{gathered}$$

Here, we can observe that the value of $f\text{'}\left(x\right)$ is positive in the range $(-2,0)\cup (2,\infty )$.

Hence it is increasing in that range.

Now, we have $-3\in \left(-\infty ,-2\right)$

$$\begin{gathered} f\text{'}(-3)=4{\left(-3\right)}^3-16\left(-3\right) \\ =-108+48 \\ =-60<0 \end{gathered}$$

and, $1\in \left(0,2\right)$

$$\begin{gathered} f\text{'}\left(1\right)=4{\left(1\right)}^3-16\left(1\right) \\ =4-16 \\ =-8<0 \end{gathered}$$

Here, we can observe that the value of $f\text{'}\left(x\right)$ is negative in the range $(-\infty ,-2)\cup (0,2)$.

Hence it is decreasing in that range.

Thus, the function $f\left(x\right)=x^4-8x^2$ is increasing in the interval $(-2,0)\cup (2,\infty )$, and decreasing in the interval $(-\infty ,-2)\cup (0,2)$.