Find the maximum and the minimum values of f(x)=3x^4-8x^3+6x^2 on [-1/2,1/2] by finding the points where the derivative is 0 and compering with the values at the end point?

1 Answer
Apr 11, 2018

f'(x)=0

0=3x^4-8x^3+6x^2

0=(x^2)(3x^2-8x+6)

use the quadratic formula to find that 3x^2-8x+6 have factors .610 and -3.277

x= 0, .610, -3.277

since range is [-.5, .5], x=0

find max/min value, plug x=0 into f(x)

compare answer with f(-.5) and f(.5)

you should be able to do the rest youself