# Math

posted by on .

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.

Also, the units are in radians.

• Math - ,

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)*2x+2x
so solve for
ln(2)*2x+2x = 0
the solution of x will be the position of minimum, as shown in graphics.
I get xmin=-0.2845 for the position of minimum, and f(xmin)=0.90197

So the fMin that you got corresponds to the value of x at the point of minimum.

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