Find-maximum-value-of-f-x-16-x-2-x-2-9-

Question Number 134285 by bramlexs22 last updated on 02/Mar/21

Find maximum value of

f (x) = \sqrt{(16 - x^{2}) (x^{2} - 9)}

Answered by EDWIN88 last updated on 02/Mar/21

let g (x) = (16 - x^{2}) (x^{2} - 9) \Rightarrow \frac{dg (x)}{dx} = - 2 x (x^{2} - 9) + 2 x (16 - x^{2}) = 0

- 2 x {x^{2} - 9 - (16 - x^{2})} = 0

- 2 x ({2 x}^{2} - 25) = 0 \to {\begin{cases} x = 0 \\ x = \pm \frac{5}{\sqrt{2}} \end{cases}

\frac{d^{2} g (x)}{{dx}^{2}} = - {12 x}^{2} + 60 < 0 for x = \pm \frac{5}{\sqrt{2}}

it follow that g (x) maximum at x = \pm \frac{5}{\sqrt{2}}

so f {(x)}_{\max} = \sqrt{(16 - \frac{25}{2}) (\frac{25}{2} - 9)}

= \sqrt{\frac{7}{2} \times \frac{7}{2}} = \frac{7}{2} .

Commented by bramlexs22 last updated on 02/Mar/21

Answered by mr W last updated on 02/Mar/21

\frac{a + b}{2} ⩾ \sqrt{a b}

f (x) = \sqrt{(16 - x^{2}) (x^{2} - 9)}

⩽ \frac{1}{2} (16 - x^{2} + x^{2} - 9) = \frac{7}{2}

i . e . m a x i m u m = \frac{7}{2}

Commented by bramlexs22 last updated on 02/Mar/21

AM - GM