Question
Find the interval for which $f(x)=x^4-2 x^2$ is increasing or decreasing.
✓
Answer
Given $f(x)=x^4-2 x^2$
So,$f^{\prime}(x)=4 x^3-4 x$
$f^{\prime}(x)=4 x\left(x^2-1\right)$
$f^{\prime}(x)=4 x(x+1)(x-1)$
For $f(x)$ is increasing $f^{\prime}(x) > 0$
So, $4 x(x+1)(x-1)>0$
$\Rightarrow x(x+1)(x-1) > 0$
$\Rightarrow -1< x <0 \text { or } x>1$
$\Rightarrow x \in(-1,0) \cup(1, \infty) $
So, $f(x)$ is increasing in interval $(-1,0) \cup(1, \infty)$.
$f(x)$ is decreasing $f^{\prime}(x)<0$
$\Rightarrow 4 x(x+1)(x-1)< 0$
$\Rightarrow x(x+1)(x-1)< 0 \therefore 4 > 0$
$\Rightarrow x< -1 \text { or } 0< x <1$
$\Rightarrow x \in(-\infty,-1) \cup(0,1)$
So, $f(x)$ is decreasing in interval $(-\infty,-1) \cup(0,1)$
So,$f^{\prime}(x)=4 x^3-4 x$
$f^{\prime}(x)=4 x\left(x^2-1\right)$
$f^{\prime}(x)=4 x(x+1)(x-1)$
For $f(x)$ is increasing $f^{\prime}(x) > 0$
So, $4 x(x+1)(x-1)>0$
$\Rightarrow x(x+1)(x-1) > 0$
$\Rightarrow -1< x <0 \text { or } x>1$
$\Rightarrow x \in(-1,0) \cup(1, \infty) $
So, $f(x)$ is increasing in interval $(-1,0) \cup(1, \infty)$.
$f(x)$ is decreasing $f^{\prime}(x)<0$
$\Rightarrow 4 x(x+1)(x-1)< 0$
$\Rightarrow x(x+1)(x-1)< 0 \therefore 4 > 0$
$\Rightarrow x< -1 \text { or } 0< x <1$
$\Rightarrow x \in(-\infty,-1) \cup(0,1)$
So, $f(x)$ is decreasing in interval $(-\infty,-1) \cup(0,1)$
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