Graceapiromanongidn
Answered

For all your questions, big or small, IDNLearn.com has the answers you need. Our platform offers comprehensive and accurate responses to help you make informed decisions on any topic.

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

We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com has the answers you need. Thank you for visiting, and we look forward to helping you again soon.