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

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The discussion centers on proving that a differentiable function f is strictly increasing on an interval I when its derivative f' is greater than 0 throughout I, except possibly at a single point where f' is non-negative. The proof involves selecting two points a and b within the interval such that a < b, leading to the conclusion that f(b) > f(a) due to the positivity of the derivative. The initial confusion regarding the implications of f' = 0 was resolved, confirming that the function remains strictly increasing despite this condition.

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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


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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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Do you have the integral and the fundamental theorem of calculus available to you, or only the derivative?
 
Ah its cool. I managed to work it out. And I am pretty convinced that it works.
Thanks anyway appreciate it.
 

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