intermediate value theorem?

Question

i'm not sure how to with e^x

f(x) e^x + x, [-1,0]

is f(x) supposed to be tween -1 and 0?

Answer 1

Did you leave out an equal sign? Is it supposed to be:

f(x) = e^x + 1 ?

Usually, this notation means that x is in the range -1 and 0, inclusive.

You are not sure how to do what?

Answer 2

I'm assuming that what you're stating is that:

Given the function: f(x) = e^x + x defined over the closed interval:
-1 < = x < = 0 verify that there is an x in the closed interval (that is find the x in the closed interval) such that the derivative at that x equals the slope of the line between the endpoints of the interval.

[ I come up with x = -.458675 on my first "gropings".]

I think the intermediate value theorem says something like this:
Given a function f(x) continuous over the closed interval [a, b] -- a domain for x -- then there exists a "c" within the interval
[a < c
[f(b) - f(a)] / [b-a] In my groping this = 1.6321

[The F'(c) above, is the derivative evaluated at x = c. You have to take the derivative first; then evaluate it at c.]

F'(c) = [ F'(x) evaluated at c ] = [ e^c + 1 ]; set this equal to the 1.6321 and solve for c.