Question

find the abs max for (-1=.5and e^-2--->=0.1 can someone explain that please...

Answer 1

Happy for another chance to help you with the same question.

I take it you're in Calculus.

Given the function f(x)= (e^x)-1-x, on the interval [-1,1], you want to determine the maximum value of this function on that interval (or, in other words, the absolute maximum of f(x) on [-1,1].

Doing this requires a 2-part process.

Part A: Find the derivative f '(x); set it = to zero, and solve for x. Calculate the associated y value from f(x).

Part B: Evaluate the function at the given endpoints, that is, at x=-1 and x=1.

Now, in this case, you'll have 3 x-values, 1 from Part A and 2 from Part B.

Make a table, quickly:

x f(x)
--- -----
-1 ?
0 ?
1 ?

WHICH f(x) value is the greatest? That POINT is your ABS MAX.

Please try these steps.

Let me know if y ou have further questions.

I'll be unavailable for the next 1/2 hour.

Answer 2

but i can get e^-2,how my teacher said e^-2?

x=0--->y=0
x=-1---->y=e^-1
x=1---->y=e-2

Answer 3

I essentially agree with these 3 values for f(x) when given x: {0, -1, 1}.

Note that this is exactly what I asked you to do in Part B, above.

Where do you need clarification or help?

Answer 4

"find the abs max for (-1<x<1) "

for \(x\in (-1,1)\) the function does not have an absolute max

Answer 5

I agree. If you're restricted to (-1,1), then x can't = -1 or 1. I'd assumed, incorrectly, that we were working with [-1,1]. Sorry.

So now let me ask whether there's (1) an absolute minimum and/or (2) a relative (or local) minimum.

Thanks for your input, Captain @Zarkon !

Answer 6

that means my teacher has mistic

Answer 7

mistake**

Answer 8

@mik333: Not at all! (Teacher has not made a mistake.)
You need to be able to recognize whether or not a given function has an absolute max or abs. min. on a given open or closed interval.
If the interval is open, as it is in this homework problem, then sorry, there is neither abs. min. nor abs. max. on that interval.

Answer 9

If the interval is closed, the abs. max, if there is one, may or may not coincide with the relative/local max. And so on.

Answer 10

ot it close interval