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

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In summary, the Mean Value Theorem can be used to prove that if f(a)=g(a) and f'(x)>g'(x) for all x>a, then f(x)>g(x) for all x>a. This can be shown by setting h(x) = f(x) - g(x), applying the MVT on the interval [a,x], and using the fact that h'(c) = f'(c) - g'(c) > 0 on the interval."
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NWeid1
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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). This is what I got:

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].

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

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

But I feel like this is wrong?!
 
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  • #2
NWeid1 said:
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). This is what I got:

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].

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

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

But I feel like this is wrong?!
The equalities that all begin with h'(c)= look very wrong. They're saying that a single number h'(c) is equal to several different numbers. That clearly can't be true. I also don't see why you conclude that h'(c)>0.
 
  • #3
Well, I know to set h(x) = f(x) - g(x) but I don't know where to go from there. lol
 
  • #4
NWeid1 said:
Well, I know to set h(x) = f(x) - g(x) but I don't know where to go from there. lol
The good news is that that's the right way to start. :smile:

Can you tell me exactly what the MVT says about h on the interval [a,x]?
 

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

1. What is the MVT (Mean Value Theorem) and how does it apply to this statement?

The MVT is a theorem in calculus that states that if a function is continuous on a closed interval and differentiable on the open interval, then there exists a point within that interval where the slope of the tangent line is equal to the average rate of change of the function. In this statement, we can use the MVT to prove that f(x) is greater than g(x) by showing that their slopes are different and therefore, there must be a point where their tangent lines intersect.

2. How does the given condition of f'(x) and g'(x) being different for all x help in proving the statement?

The condition of f'(x) and g'(x) being different for all x implies that the slopes of the tangent lines for f(x) and g(x) are also different for all x. This means that there must be a point within the interval where the slope of f(x) is greater than the slope of g(x). This point can be found using the MVT and can be used to prove that f(x) is greater than g(x) for all x within the interval.

3. Can you provide an example of how the MVT can be applied to proving the given statement?

Sure, let's say we have two functions, f(x) = x^2 and g(x) = x. We want to prove that f(x) > g(x) for all x. First, we can show that f(a) = g(a) by plugging in any value of a. Then, we can calculate the derivatives of both functions, f'(x) = 2x and g'(x) = 1. Since these derivatives are different for all x, we can use the MVT to find a point c within our interval where the slopes of the tangent lines are equal. In this case, c = 1. Therefore, at x = 1, f(x) = g(x) and f'(x) > g'(x), proving that f(x) is always greater than g(x) within the given interval.

4. Is the MVT the only way to prove the given statement?

No, the MVT is not the only way to prove the given statement. Other methods, such as using the definition of a derivative or using the first derivative test, can also be used to show that f(x) is greater than g(x) for all x within the interval. However, the MVT provides a straightforward and concise proof.

5. Can this statement be applied to all functions or are there specific requirements for it to be true?

This statement can be applied to any continuous and differentiable functions. The MVT can only be used if the functions satisfy these conditions. Additionally, the functions need to have the same value at a specific point (f(a) = g(a)) and their derivatives must be different for all x within the given interval. If these conditions are met, then the statement can be proven using the MVT.

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