# relative minimum

• April 6th 2008, 11:15 AM
andrewsx
relative minimum
Let f be the function given by f(x)= x^4/16-x^3+3x^2/8-2x. The function f has a relative minimum at x=

Thanks.

I took the derivative to get f'(x)=x^3-12x^2+3x-8 and set that equal to zero...then synthetic division??
• April 6th 2008, 11:23 AM
Moo
Hello,

Watch out, you have to solve f'(x)=0, so you can simplify by 4, but it's completely incorrect to say that f'(x)=x^3-12x^2+3x-8

f'(x)=1/4*(x^3-12x^2+3x-8)
• April 6th 2008, 11:26 AM
Mathstud28
If this is supposed to be
$\frac{x^4}{16}-x^3+\frac{3x^2}{8}-2x$...then the derivative is $f'(x)=\frac{x^3}{4}-3x^2+\frac{3x}{8}-2$ then use either, graphing calculator, newton's approximation, or use intermediate value theorem noting $f'(11)=\frac{-225}{8}$ and $f'(12)=\frac{5}{2}$