Find the minimum of g(t)=t^3-4t+8 on the interval [-1,2]
I have the derivative of g'(t)=3t^2-4.

Question

Find the minimum of g(t)=t^3-4t+8 on the interval [-1,2]
I have the derivative of g'(t)=3t^2-4.

bahrom7893 · Answer

Now set it equal to 0:
3t^2-4=0

bahrom7893 · Answer

Solve for t. What do you get?

anonymous · Answer

I get 2sqrt(1/3), -2sqrt(1/3)

bahrom7893 · Answer

Are they both included in the interval [-1, 2] ?

anonymous · Answer

no

bahrom7893 · Answer

Why not? What are the approximate values of 2*sqrt(1/3) and -2*sqrt(1/3) ?

anonymous · Answer

Ah, so is only 2*sqrt(1/3) is only one that is included. Right?

bahrom7893 · Answer

Yes. Now evalute f(x) at x=-1, x=2*sqrt(1/3), and x=2
The smallest of those is your minimum.

anonymous · Answer

so I would plug them into t?

bahrom7893 · Answer

Yea, sorry that's what I meant, f(t) at t=-1, t=2*sqrt(1/3), t=2. This is what you need basically|dw:1382487103070:dw|
That's one of the ways your graph could look like.. btw I have no idea what it looks like, this is just a random drawing.