How do you find the critical numbers of #s(t)=3t^4 + 12t^3-6t^2#?

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Answer 1 · 2018-03-11T17:09:37

#t=0# and #t=(-3+-sqrt(13))/2#

The critical points of a function is where the function's derivative is zero or undefined.

We begin by finding the derivative. We can do this using the power rule:

#d/dt(t^n)=nt^(n-1)#

#s'(t)=12t^3+36t^2-12t#

The function is defined for all real numbers, so we won't find any critical points that way, but we can solve for the zeroes of the function:

#12t^3+36t^2-12t=0#

#12t(t^2+3t-1)=0#

Using the zero factor principle, we see that #t=0# is a solution. We can solve for when the quadratic factor equals zero using the quadratic formula:

#t=(-3+-sqrt(9+4))/2=(-3+-sqrt(13))/2#