Question

find the maximum between -1 and 1 f0r the function (e^x)-x-1

Answer 1

In order to solve this problem, we have to know the definition of "maximum" and know what are the sufficient conditions for a function to have it, so, let's begin by defining "maximum":
F(x) present maximum if:
\[f(x) has M,on-> x=a <=> EE ^{*}_{a}/any"x"\in E ^{*}_{a} f(x)<f(a)\]
Let me read that for you:
"the function of x has a maximum in the point `a´if there exists a enviroment in that point such that for all the x values belonging to that point, f(x) is lower than f(a)"
By that, I mean, that in the function we will define the "maximum as the value in an interval that no value of x will surpass".
Now I'll show a theorem wich will tell us how to find the maximum of a function:


Theorem:
"if a function has a derivative of value zero, then it has a relative extrema in such point"
H) f has a relative extrema in the point x=a
f is differentiable in x=a

T) F'(a)=0

in order to prove this theorem, I'll have in consideration two situations F'(a) can take, when f'(a)>0 and when f'(a)<0.
\[if: f'(a)>0\]
Then by definition of a puntual increase then f is increasing at x=a. This is absurd, because it cannot be growing if by definition f(x)<f(a).
\[if: f'(a)<0\]
by definition of puntual decrease, we say that f is decreasing at that point, but this is absurd because no value is lower than the minimum.
sw we conclude that:
\[f'(a)=0\]
That proves our theorem.

Answer 2

my teacher said to check the mix we have tow value e^-1 ---->=.5and e^-2--->=0.1
the the maximum will be 0.5

Answer 3

Yeah, it's a way to see it. But rigurously talking, f has no maximum or minimum in that interval but you can speculate, because it's unvertain because of the form the function has.

Answer 4

so,what the answer,i do not know how he get e^-2???