Wednesday

April 1, 2015

April 1, 2015

Posted by **Ifi** on Sunday, October 28, 2012 at 12:00am.

I think the question is wrong bcus it is impossible to to have 2 unknown to find the relative max or min...right? Sorry if im wrong.

So i change it to f(x)=2x^4-2x^2-9

so for relative max is x = 0 and relative minimum is x= +/- 0.7071. Plz recheck back the answer.

- Math -
**Steve**, Sunday, October 28, 2012 at 12:39amf' = 8x^3 - 4x = 4x(2x^2-1)

you are correct

**Answer this Question**

**Related Questions**

Calculus (pleas help!!!) - Find the absolute maximum and absolute minimum ...

Calculus (pleas help!!!) - Find the absolute maximum and absolute minimum values...

Calculus - Find any absolute maximum and minimum and local maximum and minimum ...

Calculus - Find the absolute maximum and absolute minimum values of the function...

algebra - 1. How to find the numbers, if there are any, at which f has a ...

Please check my Calculus - 1. Find all points of inflection: f(x)=1/12x^4-2x^2+...

Calculus - Find the values of x that give relative extrema for the function f(x...

Caluclus - Find the absolute maximum and absolute minimum of f on the interval...

Calculus (help steve) - f(x)=−8x^3+6ax^2−3bx+4 has a local minimum ...

Math-Graphs - graph g(x)=4(x^3)-24x+9 on a calulator and estimate the local ...