Sketch the graph of y=e-x and y=lnx <No problem.>
Let f(x)=e-x- ln x and use the intermediate value theorem to show that f(x)=0 for the interval n<x<n+1 where n and n+1 are consecutive integers to be determined. If f(x0)=0,give a reason why would you expect the value of x to be less than 1.5 and verify that this is indeed true.

Thanks for help!

Apply the IVT to f(x), the closed interval [n, n + 1] and the value 0. Try different values of n. Are you allowed to use a calculator to find e^x and ln(x) for various x?

Yes, calculator is allowed. Emmm... Is it means that the value of n and n+1 can be any number but 0 must in the interval of these numbers? Or?

Originally Posted by auxianlovekakashi
Is it means that the value of n and n+1 can be any number but 0 must in the interval of these numbers? Or?
Applying the IVT to the function f(x) on the interval [n, n + 1] and the number 0 requires that 0 ∈ [f(n), f(n + 1)] or 0 ∈ [f(n + 1), f(n)] (depending on which of f(n) or f(n + 1) is greater), but not 0 ∈ [n, n + 1].

I recommend the following. Study the statement of version I of the IVT. Try different values of n = 0, 1, 2, ... For each value of n, do the function f(x), the interval [n, n + 1] and the number u = 0 fulfill the assumptions of the IVT? If they do, what does the conclusion say?

I think you have stated the problem incorrectly. The way you have written it you need to find a integer, n, such that f(x)= 0 for all x between n and n+1. What you should have is a value of n such that f(x)= 0 for some x between n and n+ 1. Since you already have graphs, look where those graphs intersect and choose n to be the largest integer less than that x. You can then "use the intermediate value theorem" to show that f(x) must be 0 between n and n+1 by evaluating f(n) and f(n+1) and seeing that one is negative and the other is positive.

What is the "intermediate value theorem"?

Ok. Thanks a lot for helping!!