Question

$f(x)= {x}^{2} + \frac{16}{ {x}^{2} } \: for \: maxima \: and \: minima$
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Answer 1

Step-by-step explanation:

$\blue{ \bf{ \underline {QUESTION : - }}}$

$\sf{ f(x)= {x}^{2} + \frac{16}{ {x}^{2} } \: for \: maxima \: and \: minima}$

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$\boxed{ \huge{ \bold{Given}}}$

$\bold{ \it{f(x)= {x}^{2} + \frac{16}{ {x}^{2} }}}$

$\huge\boxed{ \bold{to \: find}}$

• f(x) minimum value

$\star{ \pink{ \underline{ \underline{Ｓｏｌｕｔｉｏｎ : - }}}}$

$\large{ \sf{f(x) = {x}^{2} + \frac{16}{ {x}^{2} }}}$

$\large{ \sf{f'(x) = 2x - \frac{32}{ {x}^{3} }}}$

$\large{ \sf{ \implies{f'(x) = 0 = 2x - \frac{32}{ {x}^{3} }}}}$

$\large{ \sf{ {x}^{4} = 32 = {2}^{4} = 16=>x ±2}}$

$\large{ \sf{f(x) = {2}^{2} + \frac{16}{ {2}^{2} } = 4+ 4 = 8}}$

$\large{ \sf{f''(x) = 2 - \frac{96}{ {x}^{4} } > 0}}$

$\therefore{ \green{ \bf{f(x) \: will \: have \: minium}}}$

$\boxed{ \red{ \mathfrak{so \: {the \: minium \: value \: of \: f(x) = { \boxed{ \purple{8}}}}}}}$

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