Friday

March 6, 2015

March 6, 2015

Posted by **moises** on Sunday, April 28, 2013 at 3:40am.

- calculus -
**Steve**, Sunday, April 28, 2013 at 6:10amcheck f(-2) and f(2).

f' = -4x^3 + 6x^2

= -2x^2(2x-3)

so there are local extrema at x = 3/2

Not at x=0, since that's a double root of f'=0.

Now pick the max/min values of f at -2,3/2,2

- calculus -
**Reiny**, Sunday, April 28, 2013 at 7:09amf ' (x) = - 4x^3 + 6x^2

= 0 for max/min

x^2(-4x + 6) = 0

x = 0 or x = 3/2

f(0) = 5

f(3/2) = -81/64 + 54/8 + 5 = 671/64 = appr 10.48

f(-2) = - 16 + 16 + 5 = 5

f(2) = -16 - 16 + 5 = -27

so the absolute max is 671/64 and the min is -27

**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 the absolute maximum and absolute minimum values of the function...

calculus - 2) Let g(s)= t(4−t)^1/2 on the interval [0,2]. Find the ...

Calculus - Let f(t)=t\sqrt{4-t} on the interval [-1,3]. Find the absolute ...

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

Calculus - Find the absolute extrema of the function on the interval [2, 9]. (...

calculus - Let g(x)=(4x)/(x^2+1) on the interval [-4,0]. Find the absolute ...

calculus - Let g(s)=1/(s-2) on the interval [0,1]. Find the absolute maximum and...

calculus - Let f(x)=-x^2+3x on the interval [1,3]. Find the absolute maximum and...