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Answered

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Problem 1.9. Find the relative ertrema of the following function. Determine whether each
ertremum is a marimum or a minimum.
f(x) =
[tex] \frac{ {x}^{2} }{ {x}^{4} + 16 } [/tex]
​

Sagot :

Really226idn

Answer:

Min: 0

Max: 0.125

Step-by-step explanation:

[tex]\frac{d}{dx} \frac{x^2}{x^4+16} =0\\\frac{-(2x)(x^4-16)}{(x^4+16)^2} =0\\x=0, 2, -2\\f(0)=0\\f(2)=f(-2)=0.125\\\frac{d^2}{dx^2} f(x)=\frac{6x^8-384x^4+512}{(16+x^4)^3} \\f''(0)=0.125>0\\f''(2)=f''(-2)=-0.125[/tex]

Min: 0

Max: 0.125

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