hi for any x>x0 , can we say that if f(x) <= g(x) then, f'(x) < g'(x) ?
I doubt it.
What if f(x)=2 and g(x)=4, both have the same slope.
so what about if we say f'(x) < = g'(x) I need it to solve something else really need this look this is my explanation if f < g suppose it and... lets suppose that slope of f is greater than g then from one certain x to upper x emmm.... f will croos g and goes upper than g right? so... if f always be smaller than g so it has to has smaller slope am i rught?
Not necessarily.
why?
If f = g, then f ' =g '
so
will it be true if i say ..... f' < = g' ?
i think it is true then
emmm... i think the functions must be countinuse... we need this condition
Not just continuous, but smooth, otherwise you can't find the derivative.
oh.... yes... thats sign of my confused mind... it is nessesery , thanks
I'm still not sure that it is true to say, "If f(x) ≤ g(x) then f '(x) ≤ g '(x) for all x." I think f '(x) can locally be larger than g '(x) in some points, but overall f(x) will still be less than g(x).
f' is slope of f ,let me say my reasons in a better way... down of g we have f if slope of f = slope of g then they keep their distance for all x ... forever... if slope of f < slope of g : their distance becomes more and more as x ---> large values if slope of f > slope of g : their distannce become less until finally they meet anf then f goes upper than g which is counter of f<g . so f ' < = g '
So you're saying that f ' is always greater than g ' at every point and not just at particular points?
not exactly i say for x > x0 .... from somewhere to... infinity
what u said cant happen because of countinuse condition of functions
What if f '(x) is less than g '(x) in some places, but is greater in others?
i think it cant happen, let me think and give a good reason
I could just draw a picture as an example.
ohhhhhhhhhhhhhhhhhhhhhhhh, u r right. i am sorry
yes i did too, u r right
U might be able to do it over a specified interval, that's about it, though...
Yep, you can state something similar, but you have to be careful about how you word the conditions.
emmm... right... thanks CliffSedge... u helped me...thank youuuuuuuuuuu others
i wanted to use this to prove something else , now i understood it is not true and have to find other way, u made i stop waisting time, thanks, later :)
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