Math
posted by Monique .
Using the TI89 Titanium calculator:
Find the largest and smallest values of each of the following functions on their given intervals.
(I'll show you one)
f(x) = 2^x + x^2 [4, 1]
I put in the function on the Y= screen as shown, then 4 < x < 1 (it's actually the less than or equal to sign)
It gives me a little more than half of the parabola, but I don't know where to go from there. I tried punching in the same as above using fMin( and fMax( on the Home screen and got .285 for the minimum and 4 for the maximum, but I'm not sure if this is right.
Help please? Thank you!
Also, the units are in radians.

Does the TI89 Titanium have a derivative function?
If it does, solve for
f'(x)=0, or do it by hand,
f'(x)
=d(2^x+x^2)/dx
= ln(2)*2^{x}+2x
so solve for
ln(2)*2^{x}+2x = 0
the solution of x will be the position of minimum, as shown in graphics.
I get x_{min}=0.2845 for the position of minimum, and f(x_{min})=0.90197
So the fMin that you got corresponds to the value of x at the point of minimum.