Proof involving the mean value theorem and derivatives

In summary, for any non-negative real number \mu and any real number s greater than or equal to 1, it can be proven that (1+s)^{\mu} is greater than or equal to 1 + s^{\mu}. This can be done algebraically, without using the binomial theorem, by defining \mu with the help of logarithms.
  • #1
imurme8
46
0

Homework Statement



For [itex]\mu\geq 0, s\geq 1,[/itex] prove that [itex](1+s)^{\mu}\geq 1 + s^{\mu}[/itex]



Homework Equations





The Attempt at a Solution



I have written a proof involving the mean value theorem and derivatives, but there must be a simpler way! I think this should be done purely algebraically. Instructor insists that [itex]\mu[/itex] is an arbitrary non-negative real number, not just a rational or an integer. So to define it we need log, etc. But I believe there is a solution that does not go into this...
 
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  • #2


Forgot to say, no binomial theorem allowed
 
  • #3


Whoops, meant [itex]\mu \geq 1[/itex] and [itex]s\geq 0[/itex], not the other way around. However, they are both reals.
 
  • #4


imurme8 said:

Homework Statement



For [itex]\mu\geq 0, s\geq 1,[/itex] prove that [itex](1+s)^{\mu}\geq 1 + s^{\mu}[/itex]
What does this question have to do with the subject "ddd" of the thread?
 
  • #5


Whoops, meant [itex]\mu \geq 1[/itex] and [itex]s\geq 0[/itex], not the other way around. However, they are both reals.
 
  • #6


LCKurtz said:
What does this question have to do with the subject "ddd" of the thread?
I apologize, I forgot to enter a thread name that makes sense. If we can delete this one and repost it, I'd be happy to.
 
  • #7


imurme8 said:
I apologize, I forgot to enter a thread name that makes sense. If we can delete this one and repost it, I'd be happy to.

(I changed the thread title for you.)
 

1. What is the mean value theorem?

The mean value theorem states that for a continuous and differentiable function on an interval, there exists at least one point in that interval where the slope of the tangent line is equal to the slope of the secant line connecting the endpoints of the interval.

2. How is the mean value theorem used in proofs?

The mean value theorem is often used in proofs to show that a function has a certain property or to prove the existence of a value. It allows us to connect the behavior of a function at two points with the behavior of its derivative at a single point.

3. What is the role of derivatives in mean value theorem proofs?

Derivatives play a crucial role in mean value theorem proofs as they represent the instantaneous rate of change of a function at a specific point. They allow us to compare the slope of the tangent line at a point with the slope of the secant line connecting two points on the function.

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. This means that the function must be continuous on the interval and have a defined derivative at every point in the interval.

5. Can the mean value theorem be used to find the exact value of a function?

No, the mean value theorem only guarantees the existence of a point where the slope of the tangent line is equal to the slope of the secant line. It does not provide the exact value of this point or the function itself.

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