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Find the intervals in which the function \(f\) given by $f(x) = x^3 + \large\frac{1}{x^3}$$, x \neq 0$ is $(i) \;increasing\qquad (ii) \;decreasing.$

1 Answer

  • $a^2-b^2=(a+b)(a-b)$
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
We have $f(x)=x^3+\large\frac{1}{x^3}$$x\neq 0$
For $f(x)$ is an increasing function of $x$,
(i.e) $3\big(x^2-\large\frac{1}{x^4}$$\big)>0$
$\Rightarrow x^6-1>0\Rightarrow (x^3-1)(x^3+1)>0$
$x^3-1>0$ and $x^3+1>0$
$\Rightarrow x^3>1$ & $x^3>-1$
$\Rightarrow x>1$ & $x>-1$
$\Rightarrow x^3-1>0$ and $x^3+1<0$
$x^3<1$ and $x^3<-1$
$x<1$ and $x<-1$
$\Rightarrow x<-1$
Hence $f(x)$ is increasing when $x<-1$ and $x>1$
Step 2:
For $f(x)$ to be decreasing function of $x$
$f'(x)<0$ i.e $3\big(x^2-\large\frac{1}{x^4}\big)$$<0$
$\Rightarrow x^2-\large\frac{1}{x^4}$$<0\;\;\;x^6-1<0$
$\Rightarrow (x^3-1)(x^3+1) <0$
Step 3:
Let (i)$x^3-1>0$and $x^3+1<0$
$\quad (ii)x^3-1<0$and $x^3+1>0$
(i) $x^3>1$ and $x^3<-1$ or $x>1$ and $x<-1$ not possible.
(ii)$x^3<1$ and $x^3>-1$ (or) $x<1$ and $x>-1$
(i.e)-1< $x$ <1
Hence $f(x)$ is decreasing when -1 < $x$< 1
answered Aug 10, 2013 by sreemathi.v

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