Question

f(t)=t^4-8t^2 on the interval [-3,3]

find the critical points and the max and min

Answer 1

You can use Calculus to find the crititical points.

Answer 2

You just need to get the first derivative and see where it equals to zero. \[\frac{df(t)}{dt}=0\]

Answer 3

In this case \[f'(t)=4t^3-8t=t(4t^2-8)=0\]

Answer 4

So, there is critical point at \[t = -\sqrt{2},0,\sqrt{2}\]