# Critical Numbers

• May 18th 2010, 01:06 PM
sheva2291
Critical Numbers
Find the critical numbers of the function. (Round your answers to three decimal places.)
s(t) = 3t4 + 4t3 − 6t2
• May 18th 2010, 01:21 PM
Quote:

Originally Posted by sheva2291
Find the critical numbers of the function. (Round your answers to three decimal places.)
s(t) = 3t4 + 4t3 − 6t2

EDIT: Hold on!
• May 18th 2010, 01:23 PM
Set $s'(t)=0$.
You get (I assume by t2 you mean $t^2$ for example) $12t^3 + 12t^2 - 12t = 0$.
$t=0$ is an obvious solution...
Divide through by $12t$ to get
$t^2 + t - 1 = 0$
Solve this using standard $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ formula