Using the Mean Value Theorem to Prove Inequality for e^x and 1 + x

[VeeEight]

The following two questions are practice problems that I have been stuck on.

Homework Statement

Use the Mean Value Theorem to show that e^x > 1 + x for all x > 0

Homework Equations

Mean Value Theorem: If f: [a,b] to R is continuous on [a,b] and differentiable on (a,b) then there exists a point c in (a,b) where f(c) = $$f(b) - f(a)/b - a$$

The Attempt at a Solution

I can do the question by induction so I was thinking about first showing the inequality is true for an x > 0, and then using the Mean Value Theorem in the second step to show that this implies ee^x > x + 2 for all x > 0. The only thing is that I don't know how to use the Mean Value Theorem in this situation - I've tried a few random cases but I can't think of what my interval should be in order to get the desired condition.

The second problem:

Homework Statement

Suppose f'(0) exists and f(x + y) = f(x)f(y) for all x and y. Prove that f' exists for all x.

The Attempt at a Solution

Here are some things I gathered from the given information.

f'(0) exists implies that the limit as x approaches 0 of $$f(x) - f(0)/x$$ exists.

f'(x + y) = f'(x)f(y) + f'(y)f(x)
so f'(0) = f(x - x) = f'(x)f(-x) + f'(-x)f(x)
and so f'(x)f(-x) + f'(-x)f(x) = the limit as x approaches 0 of $$f(x) - f(0)/x$$

I'm not sure if I am just going down the wrong path here since I tried to rearrange the above equation so it can look better but I got nowhere.

 

[Feldoh]

f'(c) is the going to be somewhere on an interval [a,b].

For e^x this means that e^c = the average slope on the interval [a,b]
For 1+x this means the f'(c) = 1 = the average slope on the interval [a,b]

We know the e^c > 1 where c > 0