Proof: f is Strictly Increasing in an Interval with f' > 0 | Homework Analysis

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In summary, if f is a differentiable function with a positive derivative throughout an interval except possibly at a single point where the derivative is non-negative, then f is strictly increasing on that interval. This can be shown using the definition of the derivative and the fact that the derivative is positive. The case where the derivative is equal to zero can be handled using the fundamental theorem of calculus.
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Homework Statement


If f is differentiable in an interval I and f' >0 throughout I, except possibly at a single point where f' >=0 then f is stictly incresing on I


Homework Equations





The Attempt at a Solution



Ok what I have is I let f'(x) >0. I let a and b two points in the interval with a<b. then for some x in (a,b) with
F'x= (f(b)-f(a))/b-a
but f'(x)>0 for all x in (a,b) so
(f(b)-f(a))/(b-a) >0

since b-a>0 it follows that f(b)>f(a)


What you can see I have proved that it is incresing in the interval but I am not sure what to do when f'=0 any help would be much appreciated as I have been told that it is not fully correct.
 
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  • #2
Do you have the integral and the fundamental theorem of calculus available to you, or only the derivative?
 
  • #3
Ah its cool. I managed to work it out. And I am pretty convinced that it works.
Thanks anyway appreciate it.
 

FAQ: Proof: f is Strictly Increasing in an Interval with f' > 0 | Homework Analysis

What does it mean for a function to be strictly increasing in an interval?

A function is considered strictly increasing in an interval if its output values consistently increase as the input values increase within that interval. In other words, as the input values move along the x-axis, the corresponding output values move only in the positive direction along the y-axis.

What is the significance of f' > 0 in this statement?

The notation f' > 0 refers to the derivative of the function f, which represents the rate of change of the function at a particular point. A positive value for the derivative indicates that the function is increasing at that point, which is necessary for a function to be strictly increasing in an interval.

How is the proof of a function being strictly increasing in an interval with f' > 0 different from other types of proofs?

The proof of a function being strictly increasing in an interval with f' > 0 requires the use of calculus concepts such as derivatives and the Mean Value Theorem. This is because the concept of strict increasing is closely related to the rate of change of a function, which is represented by the derivative. Other types of proofs may involve different mathematical concepts and techniques depending on the specific statement being proven.

Why is it important to prove that a function is strictly increasing in an interval with f' > 0?

Proving that a function is strictly increasing in an interval with f' > 0 is important because it guarantees that the function will consistently increase within that interval. This information can be useful in various applications, such as optimizing a function or determining the behavior of a system over time. It also provides a stronger understanding of the behavior of the function and its relationship to its derivative.

Can a function be strictly increasing in an interval with f' > 0 if the derivative is zero at some points within that interval?

No, a function cannot be strictly increasing in an interval with f' > 0 if the derivative is zero at any point within that interval. This is because a derivative of zero indicates that the function has a constant rate of change, and therefore there is no consistent increase in the output values within that interval.

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