Mean Value Theorem Question

In summary: I don't know how to graph that. :(In summary, the function G(x) = x^2 - e^{\frac{1}{1+x}} assumes a value of 0 for exactly one real number x such that 0 < x < 2.
  • #1
calcnd
20
0
Use the Intermediate Value Theorem and/or the Mean Value Theorem and/or properties of [tex]G'(x)[/tex] to show that the function [tex]G(x) = x^2 - e^{\frac{1}{1+x}}[/tex] assumes a value of 0 for exactly one real number x such that 0 < x < 2 . Hint: You may assume that [tex]e^{\frac{1}{3}} < 2 [/tex].

So I'm completely lost.

Here's the first thing I tried:

[tex]G(x) = 0[/tex]

[tex] 0 = x^2 - e^{\frac{1}{1+x}} [/tex]

[tex]e^{\frac{1}{1+x}} = x^2[/tex]

[tex] lne^{\frac{1}{1+x}} = lnx^2[/tex]

[tex] \frac{1}{1+x} = 2lnx [/tex]

[tex] 1 = 2(1+x)lnx [/tex]

[tex] \frac{1}{2} = (1+x)lnx[/tex]

Which isn't quite getting me anywhere.

And I'm not sure how the mean value theorem is going to help me out much more since:

[tex]G'(c) = \frac{G(2) - G(0)}{2-0}[/tex]

[tex]2c + (1+c)e^{\frac{1}{1+c}} = \frac{[2^2-e^{\frac{1}{3}}] - [0^2 - e]}{2}[/tex]

[tex] 4c + (2+2c)e^{\frac{1}{1+c}} = 4 - e^{\frac{1}{3}} + e [/tex]

Which isn't going to simply easily.

Oy vey...

Any help or hints would be appreciated.

Edit: There, I think I finally got it to render correctly.
 
Last edited:
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  • #2
Intermediate value theorem: G(0)<0, G(2)>0, so there is at least one root between. Now, how can you show it's going to be unique using G'(x)?
 
  • #3
Hrm, that's what I'm not sure of.

I know the x value is within the interval, due to the intermediate value theorem, as you stated.

I could suggest that I solve [tex]G'(x)=0[/tex], but I could imagine that would be useless as there's no reason to think that [tex]G(c)=0[/tex] is a critical point.

:S
 
  • #4
Look at [tex] G'(x)[/tex] and decide what its properties are for 0<x<2. Then decide what that tells you about what [tex]G(x)[/tex] is doing on (0,2).
 
  • #5
Heh... I noticed I did my derivative entirely wrong. I guess that's why you shouldn't use the computer when you're tired.

Anyways,

[tex]G(0)=0^2-e^(\frac{1}{1+0})[/tex]
[tex]=-e[/tex]

Which is < 0

[tex]G(2)=2^2-e^{\frac{1}{1+2}}[/tex]
[tex]=4-e^{\frac{1}{3}} [/tex]

Which is > 0

Meaning [tex]G(x)=0[/tex] is contained somewhere between x=0 and x=2.

[tex]G'(x)=2x + {e^{\frac{1}{1+x}}\over{(1+x)^2}}[/tex]

But I'm not having an easy time trying to solve this equal to zero to find my critical numbers. It seems relatively safe to assume that [tex]G(x)[/tex] is increasing on this interval.
 
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  • #6
Can you see whether G'(x) is positive or negative for positive x?
 
  • #7
I can. But how does that help me determine the value of X at which G(X) equals zero? :(
 
  • #8
Try graphing it.
 

Related to Mean Value Theorem Question

1. What is the Mean Value Theorem?

The Mean Value Theorem is a fundamental theorem in calculus that states that for a continuous and differentiable function on a closed interval, there exists at least one point where the slope of the tangent line is equal to the average rate of change of the function over that interval.

2. How is the Mean Value Theorem used in calculus?

The Mean Value Theorem is used in calculus to prove the existence of critical points, to solve optimization problems, and to prove other theorems such as the Fundamental Theorem of Calculus.

3. What are the conditions for the Mean Value Theorem to hold?

The conditions for the Mean Value Theorem to hold are that the function must be continuous on a closed interval and differentiable on the open interval, meaning that the derivative of the function must exist at every point within the interval.

4. Can the Mean Value Theorem be applied to all functions?

No, the Mean Value Theorem can only be applied to continuous and differentiable functions. If a function is not continuous or differentiable, the theorem does not hold.

5. How is the Mean Value Theorem related to the concept of average?

The Mean Value Theorem is related to the concept of average because it states that there exists at least one point within a given interval where the instantaneous rate of change is equal to the average rate of change. This point can be seen as the "average" point where the function behaves in a way that is representative of the entire interval.

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