Question
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.
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.
Answers
MathMate
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<sup>x</sup>+2x
so solve for
ln(2)*2<sup>x</sup>+2x = 0
the solution of x will be the position of minimum, as shown in graphics.
I get x<sub>min</sub>=-0.2845 for the position of minimum, and f(x<sub>min</sub>)=0.90197
So the fMin that you got corresponds to the value of x at the point of minimum.
If it does, solve for
f'(x)=0, or do it by hand,
f'(x)
=d(2^x+x^2)/dx
= ln(2)*2<sup>x</sup>+2x
so solve for
ln(2)*2<sup>x</sup>+2x = 0
the solution of x will be the position of minimum, as shown in graphics.
I get x<sub>min</sub>=-0.2845 for the position of minimum, and f(x<sub>min</sub>)=0.90197
So the fMin that you got corresponds to the value of x at the point of minimum.