Posted by **Monique** on Wednesday, June 24, 2009 at 4:27pm.

Using the TI-89 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.

- Math -
**MathMate**, Wednesday, June 24, 2009 at 6:15pm
Does the TI-89 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.

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