The Student Room Group

[C2] Differentiation - Increasing function

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Original post by TenOfThem
I agree that there seems to be an unusual element to this definition ... but it has nothing to do with f(x)=0


IF f(x1) = f(x2) then we have a horizontal line, of course it may be a very very short horizontal line. Is this not the same as saying the gradient is zero.
Original post by steve2005
IF f(x1) = f(x2) then we have a horizontal line, of course it may be a very very short horizontal line. Is this not the same as saying the gradient is zero.


yes

which is why I asked if you meant f'(x) rather than f(x)

Ah ... I see that you have edited the original comment
(edited 12 years ago)
Original post by steve2005
I think the OP has a point. Obviously, if f'(x) is increasing then it is an increasing function, and if f'(x) is decreasing it is a decreasing function.

The problem is that when f'(x) is zero it can be classed as an increasing function and a decreasing function. The problem seems that f'(x) equals zero comes in both camps.


This is why I gave 2 responses to his question

One response said "no the 2 bits of your definition are not the same"

Then the second ... the one you quoted ... ignored his odd definition and referred him back to the definition that people had been giving him throughout the thread
Reply 23
Original post by TenOfThem
This is why I gave 2 responses to his question

One response said "no the 2 bits of your definition are not the same"

Then the second ... the one you quoted ... ignored his odd definition and referred him back to the definition that people had been giving him throughout the thread


Is the rule that OP gave completely correct or wrong?

I know that for an increasing function, f(x2)>f(x1) f(x_2) > f(x_1) , when x2>x1 x_2 > x_1

But does the same applies for f(x2)=f(x1) f(x_2)=f(x_1), this means zero gradient so the function is neither increasing nor decreasing.

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