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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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The maximum value of $\Large\frac{1}{x}$ is:


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  • To obtain the absolute maxima or minima for the function $f(x)$
  • (i) Find $f'(x)$ and put $f'(x)=0$
  • (ii) Obtain the points from $f'(x)=0$
  • (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1
$y= \bigg( \large\frac{1}{x} \bigg)^x$
take log on both sides.
$ \log y=x\: \log \bigg( \large\frac{1}{x} \bigg)$
differentiating w.r.t $x$ on both sides we get,
$ \large\frac{1}{y}. \large\frac{dy}{dx}=x. \bigg( \large\frac{1}{\large\frac{1}{x}}\bigg).\bigg( -\large\frac{1}{x^2} \bigg)+ \log \bigg( \large\frac{1}{x} \bigg).1$
$ \large\frac{dy}{dx}=y \bigg( \log \bigg( \large\frac{1}{x} \bigg)-1 \bigg)$
$ =\bigg( \large\frac{1}{x} \bigg)^x \bigg[ \log \bigg( \large\frac{1}{x} \bigg)-1 \bigg]$
when $ \large\frac{dy}{dx}=0$
$ \Rightarrow \bigg( \large\frac{1}{x} \bigg)^x \bigg[ \log \bigg( \large\frac{1}{x} \bigg)-1 \bigg]$
But $\bigg( \large\frac{1}{x} \bigg)^x \neq 0$
Hence $\Rightarrow \log_e \large\frac{1}{x}-1=0$
$ \Rightarrow \log_e \large\frac{1}{x}=1$
$ e^1=\large\frac{1}{x}$
$ \therefore x=\large\frac{1}{e}$
At $f'(x)=0 \: at\: x=\large\frac{1}{e}$
$ f''(x) = \bigg[ -\large\frac{1}{x}.y \bigg]_{x=\large\frac{1}{e}}= -\large\frac{1}{x} \bigg( \large\frac{1}{x} \bigg)^x$
$ =-e.(e)^{\large\frac{1}{e}} <0$
$ \therefore \: At \: x=\large\frac{1}{e},\: y$ has maximum value and the maximum value of $f(x)\: is \: e^{\Large\frac{1}{e}}.$
answered Aug 13, 2013 by thanvigandhi_1
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