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

Question

Problem 1.9. Find the relative ertrema of the following function. Determine whether each

ertremum is a marimum or a minimum.
f(x) =

`\frac{ {x}^(2) }{ {x}^(4) + 16 }`
​

Answer 1

Min: 0

Max: 0.125

Explanation:

`(d)/(dx) (x^2)/(x^4+16) =0\\(-(2x)(x^4-16))/((x^4+16)^2) =0\\x=0, 2, -2\\f(0)=0\\f(2)=f(-2)=0.125\\(d^2)/(dx^2) f(x)=(6x^8-384x^4+512)/((16+x^4)^3) \\f''(0)=0.125>0\\f''(2)=f''(-2)=-0.125`

Min: 0

Max: 0.125