## Calculus 8th Edition

$t=0$ is the critical number
Find the critical numbers of $g(t) = t^4+t^3+t^2+1$ Differentiate and set the derivative $=0$ $g'(t) = 4t^3+3t^2+2t$ $0=t(4t^2+3t+2)$ $b^2-4ac = 9-4(4)(2) = -23$ Because the discriminant of the right hand side quadratic $<0$, it will never be $=0$. Therefore, the only value of $t$ that results in $g'(t) = 0$ is $t=0$.