Question

F(x)=e^(1-x) for 0 13 years ago

Answer 1

If the interval is open, there is no maximum or minimum. Because, note that:
\[
\frac{d}{dx}e^{(1-x)}\ne0,\; \forall x\in\mathbb{R}
\]

Answer 2

I think the the interval is closed, he was just unable to put the 'equal to' signs under the inequality symbols.

Answer 3

Since the question clearly states: "..over the prescribed closed, bounded interval."

Answer 4

it suppose to be [0 \le x \le1\]

Answer 5

Ahh, crap, I just saw the last part... if it is closed, then we have, since there is no critical point, exactly one of \(x\in \{0, 1\}\), must each be one of the maximum and minimum, evaluating the function at each \(x\) gives us the desired result.

Answer 6

So:
\[
e^{1-1}=1<e^{1-0}=e
\]So, we have a maximum at \(x=0\) and a minimum at \(x=1\).

Answer 7

By the extreme value theorem and that d/dx (e^1-x) < 0 for all x, we can say that the the maximum value over the given closed interval occurs at x = 0 and the minimum occurs at x = 1. As simple as that.

@elskydiablo

Answer 8

thank you so much

Answer 9

No problem.

Answer 10

Yep, sure thing.