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Show that the function given by \( f(x) =\large \frac{\log \: x}{x}\) has maximum at \(x = e.\)

1 Answer

  • $\big(\large\frac{u}{v}\big)'=\frac{u'v-uv'}{v^2}$
  • Maxima & Minima $f'(x)=0$
Step 1:
$f(x)=\large\frac{\log x}{x}$
Differentiating with respect to x we get
$f'(x)=\large\frac{\Large\frac{1}{x}.x-\log x.1}{x^2}$
$\qquad=\large\frac{1-\log x}{x^2}$
For maxima and minima $f'(x)=0$
$\large\frac{1-\log x}{x^2}$$=0$
$1-\log x=0$
$\log x=1$
Step 2:
On double differentiation of f(x) we get
$f''(x)=\large\frac{d}{dx}\big(\frac{1-\log x}{x^2}\big)$
$\qquad\;=\large\frac{\Large\frac{1}{x}\times x^2-(1-\log x).2x}{x^4}$
$\qquad\;=\large\frac{-x[1+2(1-\log x)]}{x^4}$
$\qquad\;=\large\frac{-(1+2-2\log x)}{x^3}$
$\qquad\;=\large\frac{-(3-2\log x)}{x^3}$
Step 3:
$f''(c)=\large\frac{-(3-2\log_e e)}{e^3}$
$\therefore f$ is maximum at $x=e$
answered Aug 9, 2013 by sreemathi.v
edited Sep 2, 2013 by sharmaaparna1

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