- f(0)=1
- |f(x) - f(y) - x^2 + y^2| ≤ (cosx-cosy)^2
Prove that g(x)=f(x) - x^2 is a constant function
>>8844764
I see you got it right this time, brainlet.
>>8844764
uwot m80
>>8844769
yeah senpai srry for that
>>8844764
pretty easy desu. Show that the derivative of f(x) is 2x using some well chosen value for y and limits.
Kek you guys are not as smart as i though
>>8846602
Already done it and it's not my homework. Also it's not quite what you said just try it nigga.
>>8846608
I actually made this and so far only 2 of my friends solved this
>>8846608
I tried it, I didn't type it until I tried, dumbass.
>>8846653
Did you actually solve it then
Rewrite the inequality as
[math] |g(x)-g(y)| \leq (cosx-cosy)^2[/math]
Say for x>y and divide by x-y
[math]\left| \frac{g(x)-g(y)}{x-y} \right| \leq \frac{(cosx-cosy)^2}{x-y}[/math]
Take the limit as [math]y\rightarrow x [/math] or w.e of both sides and then
[math]|g'(x)| \leq \lim_{x\rightarrow y}\frac{(cosx-cosy)^2}{x-y} = 0 [/math]
I am a retard and don't know what f(0) = 1 is for.
Also a limit computation that is weird
[math] \lim_{x\rightarrow y}\frac{(cosx-cosy)^2}{x-y} = \lim_{x\rightarrow y}(x-y)\frac{(cosx-cosy)^2}{(x-y)^2} =\lim_{x\rightarrow y}(x-y)\left(\lim_{x\rightarrow y}\frac{cosx-cosy}{x-y}\right)^2 = \lim_{x\rightarrow y}(x-y)\cdot (cos(x)')^2 = 0 [/math]
>>8846730
forgot to mention it doesn't matter whether y>=x or x>=y because of symmetry.
>>8846730
the problem was asking find f(x) i just made it easier by asking prove g(x) is a constant function because you have to do that first in order to find f(x) and then use f(0)=1 to find that g(x)=1 and f(x)=x^2 + 1
>>8846768
do you think if i ask find f(x) instead of prove g(x) is constant then it will be too hard for a high school student?
>>8846776
I think so. You might ask something like "find the derivative of f" first if they know a definition for derivatives and know something about limits.