Use your calculator to find the largest and smallest values of the following function in the given interval?

Question

f(x)=2^x +x^2 interval: [-4,1]

Can anyone tell me how to do this on my calculator??

Yes, I am using a TI-89...

Answer 1

First of all, enter the function in your graphing calculator, Ti-82, 83, 83plus, 84, 84 plus. If you have 86 and 89, that will be a different process.


When you graph in your calculator, choose that window for X
Xmin = -5, Xmax = 5 and Ymin = -10 and Y Max = 10
Xscale = 1 and Yscale = 1.

We need to find first if the value where the curve turns is in the interval [-4, 1].

Calculator step: Ti 82, 83, 83+ , 84, 84+

Press [2nd], Press Trace(Calc), choose Minimum. You will see your calculator might say "Left bound".

If the little dot that is flashing is in the left where you think the local minimum value is, you don't need to do anything. You will need to press Enter right away.
But if the little dot that is flashing is in the right where the local minimum is, you will need to move it to the left until it passes the minimum value. Then press enter

The calculator will say "Right bound".
You will move the cursor (or the flashing dot) to the right until you cross the minimum value. Press enter.

Note: Do not use the up and down arrow in your calculator. Use the left and right arrow that is in your calculator.

The calculator will say "Guess". Press enter. The value x is -.2845 and the y value is .9020. I round off. That means f(-.2845) = .9020. Now the value x you obtain in your calculator is in the interval [-4, 1] which is good. Now we need to calculate f(-4) and f(1) to find the largest and smallest value. With the calculator, we can do that. With the graph in your calculator, just press -4 and you will see the calculator will give you y =16.0625. Hence f(-4) = 16.0625 and press 1 in the calculator with the graph of f(x) and press enter, you will obtain 3. Therefore, f(1) = 3
Now we are going to compare f(-.2845) = .9020, f(-4) =16.0625, and f(1) = 3

Between those 3, the smallest value is .9020 and the largest value is 16.0625




I just found out that you wrote you are using Ti-89. If you are using Ti-89, that means you know calculus.


Anyway, go to your calculator Ti-89 and enter your function 2^x+ x^2 in the Y-editor, then write your function. Then press F3 to graph the function. When the graph is displayed, you wil open the Math menu by pressing F5 and choose 3 which is minimum. Press enter. The calculator will say lower bound. If you follow the same step as the explanation i gave above for ti-83 to 84 plus, you will get the answer. The "Lower bound" in Ti-89 can be referring as "Left bound" in Ti-83, 84, and The "Upper Bound" can be referring as "Right Bound". If you read the explanation above for ti-83, and 84, it will be the same thing. Except Ti-89 won't say "Guess".

Answer 2

I hope you're calucator is a TI-85 'cause that's what I'm doing mine on.

1) Select "RANGE." Change xMin to -5, xMax to 2, & yMin to -3.
2) In "y(x)=," type after y1, "2x + x²." Select "GRAPH" (M5, 2nd + F5).
3) Visually examine the graph. Assuming that x=-1 is the smallest value, press "MORE" 2x to find "EVAL" (F1). In "Eval x=," type in -1 & Enter. The result is f(-1) = -1.
4) Assuming that -4 is the largest value, type in -4 in "Eval x=," & Enter. The result is f(-4) = 8.

This took me over ten minutes 2 figure it out. I hope it helps.

Answer 3

I can only assume that you are expected to use a graphing calculator.