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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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The function $f(x)=\Large\frac{2x^2-1}{x^4}$,x>0,decreases in the interval__________.

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  • Let $f(x)$ be a function defined on $(a,b)$. If $f'(x)<0$ for all $x \in (a,b)$ except for a finite number of points, where $f'(x)=0,$ then $f(x)$ is decreasing on $(a,b)$
Step 1
$f(x)= \large\frac{2x^2-1}{x^4}$
differentiating w.r.t $x$ we get,
$f'(x)= \Large\frac{x^4(4x)-(2x^2-1).4x^3}{(x^4)^2} = \Large\frac{4x^5-8x^5+4x^3}{x^8}$
$ = \large\frac{4x^3[-x^2+1]}{x^8}$
when $f'(x)>0$
$ \Rightarrow 4x^3 \bigg[ \large\frac{-x^2+1}{x^8} \bigg]=0$
$ \Rightarrow x^2-1>0$
$ x > \pm 1$
Hence the function decreases in the interval $(-1, 1)$
answered Aug 13, 2013 by thanvigandhi_1
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