Friday

November 21, 2014

November 21, 2014

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...

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 ...

Calculus - For any constant c, define the function f_c(x)= x^3+2x^2+cx. (a) ...

Calculus - For any constant c, define the function f_c(x)= x^3+2x^2+cx. (a) ...

absolute maximum - absolute maximum of (x^2-4)/(x^2+4) the critical numbers i ...

math - Sketch f (x) = x + cos x on [-2pie,2pie ]. Find any local extrema, ...