6. Application of derivatives — ગણિત ધોરણ 12 સાયન્સ — Question

$\frac{d}{{dx}}f(x) = \frac{d}{{dx}}\left[ {4x + \frac{1}{x}} \right] $

$= 4 - \frac{1}{{{x^2}}}$

For extremum, $\frac{{dy}}{{dx}} = 0$

==> $4 - \frac{1}{{{x^2}}} = 0$

==> $x = \frac{1}{2},\, - \frac{1}{2}$

$f\;\left( {\frac{1}{2}} \right) = 4.\frac{1}{2} + \frac{1}{{1/2}}$ = $2 + 2 = 4$

$f\;\left( { - \frac{1}{2}} \right) = 4.\left( { - \frac{1}{2}} \right) + \frac{1}{{ - 1/2}} = - 2 - 2 = - 4$

Hence $f(x)$ is decreasing in interval $[1/2,\, - 1/2]$.