6. Application of derivatives — Maths STD 12 Science — Question

$f(x) = 1 + \frac{4}{{1 + x}} + (5 + x) = (6 + x) + \frac{4}{{(1 + x)}}$

==> $f'(x) = 1 - \frac{4}{{{{(1 + x)}^2}}} = 0$;

${x^2} + 2x - 3 = 0$==>$x = - 3,\;1$

Now $f''\,(x) = \frac{8}{{{{(1 + x)}^3}}}$, $f''\,( - 3) = - ve$, $f''\,(1) = + ve$

Hence minimum value at $x = 1$

$f(1) = \frac{{(5 + 1)(2 + 1)}}{{(1 + 1)}} = \frac{{6 \times 3}}{2} = 9$.