If f(a)=g(a) and f'(x)>g'(x) for all x, use MVT to prove f(x)>g(x)

[NWeid1]

1. Homework Statement
Give a graphical argument that if f(a)=g(a) and f'(x)>g'(x) for all x>a, then f(x)>g(x) for all x>a. Use the Mean Value Theorem to prove it.


2. Homework Equations



3. The Attempt at a Solution
I have sketched a graphical argument to show that f(x)>g(x). Then I applied the MVT and got

[tex]f'(c)=\frac{f(b) - f(a)}{b - a} > g'(c)=\frac{g(b) - g(a)}{b - a}[/tex]
[tex]-f'(c)(b - a) + f(b) = f(a) > -g'(c)(b - a) + g(b) = g(a)[/tex]
But now I'm stuck and don't know how to get to f(x) > g(x).

 

[Mark44]

1. Homework Statement
Give a graphical argument that if f(a)=g(a) and f'(x)>g'(x) for all x>a, then f(x)>g(x) for all x>a. Use the Mean Value Theorem to prove it.


2. Homework Equations



3. The Attempt at a Solution
I have sketched a graphical argument to show that f(x)>g(x). Then I applied the MVT and got

[tex]f'(c)=\frac{f(b) - f(a)}{b - a} > g'(c)=\frac{g(b) - g(a)}{b - a}[/tex]
[tex]-f'(c)(b - a) + f(b) = f(a) > -g'(c)(b - a) + g(b) = g(a)[/tex]
But now I'm stuck and don't know how to get to f(x) > g(x).

Much better!

Let h(x) = f(x) - g(x).

Then h'(x) = f'(x) - g'(x) > 0 for any x > a.

Now apply the MVT on the interval [a, x].

 

[NWeid1]

So.. h'(c) = (h(x) - h(a))/(x -a) ?

 

[NWeid1]

And then
[tex]h'(c) =\frac{f(x) - f(a) - g(x) + g(a)}{x - a} > 0[/tex]
[tex]h'(c) =f(x) - f(a) - g(x) + g(a) > 0[/tex]
[tex]h'(c) =f(x) - f(a) > g(x) - g(a)[/tex]
[tex]h'(c) =\frac{f(x) - f(a)}{x-a} > \frac{g(x) - g(a)}{x-a}[/tex]
==> h'(c) = f(x) > g(x)
==> f(x) > f(x) for all x>a
I feel like this is wrong?

 

[Mark44]

And then
[tex]h'(c) =\frac{f(x) - f(a) - g(x) + g(a)}{x - a} > 0[/tex]
[tex]h'(c) =f(x) - f(a) - g(x) + g(a) > 0[/tex]
[tex]h'(c) =f(x) - f(a) > g(x) - g(a)[/tex]
[tex]h'(c) =\frac{f(x) - f(a)}{x-a} > \frac{g(x) - g(a)}{x-a}[/tex]
==> h'(c) = f(x) > g(x)
==> f(x) > f(x) for all x>a
I feel like this is wrong?

For starters, h'(c) [itex]\neq[/itex] f(x), and f(x) is not larger than itself.

Since h'(x) > 0 for x > a, then h'(c) > 0 if c > a. (c comes from the MVT.)

This implies that

[tex]h'(c) =\frac{f(x) - f(a) - g(x) + g(a)}{x - a} > 0[/tex]
[tex]\Rightarrow h'(c) =\frac{f(x) - g(x) - (f(a) - g(a))}{x - a} > 0[/tex]
Can you work with this to say something about f(x) - g(x)?

 

[Fredrik]

As I said in the other thread, the equalities that start with h'(c)= don't make much sense. A single number h'(c) can't be equal to several different numbers.

Start by writing down exactly what the MVT says about the function h=f-g on the interval [a,x]. Don't just write down the equality that's a part of the statement. Write down the full statement. If you choose to use the symbol c for the the same thing as before, then the next step is to prove that h'(c)>0. When you have done that, we can start talking about how to proceed from there.

 

[NWeid1]

Mark44: I meant to say f(x) > g(x) lol. But anyways.. if
[tex]h'(c) =\frac{f(x) - f(a) - g(x) + g(a)}{x - a} > 0[/tex]
[tex]\Rightarrow h'(c) =\frac{f(x) - g(x) - (f(a) - g(a))}{x - a} > 0[/tex]

would it mean that

[tex] h'(c) = f'(c) - g'(c) > 0[/tex] ??

 

[NWeid1]

Fredrik: The MVT says that if h(x) is continuous and differentiable, then there will exist a c on [a,x] such that

[tex]h'(c) =\frac{h(x) -h(a)}{x - a}[/tex]

And now how do I prove it to be >0?

 

[Fredrik]

That's essentially correct, but there are some minor inaccuracies. I'm going to nitpick those details before I answer your question. A function is continuous at every point where it's differentiable, so it sounds a bit strange to say that h is continuous and differentiable. The MVT applies to functions on [a,x] that are differentiable on the open interval (a,x) and continuous on the closed interval [a,x]. According to Wikipedia, that "c" is a member of the open interval (a,x). I wouldn't call "h(x)" a function. h is the function, and h(x) is an element of the range of h.

The proof that h'(c)>0 is very easy. You just need to remember how you defined h, and what exactly you have assumed about f and g. The proof starts with "For all t in (a,x), h(t)=..." Can you take it from here?

Once you have proved that, you only need to think about what the things you have proved so far tell you about h(x). (Note that if I say "h(x)" I never mean the function h. I mean the value of h at x, where x is the right endpoint of the interval we're considering).