would help indeed.Hey Treb, is x a whole number? Sure would help...
I think it's pi, not n...?[maths]f'(x)=-\frac{1}{x^2}+\frac{1}{n}\frac{2nx^2\sin(nx)\cos(nx)-2x\sin^2(nx)}{x^4}\\=-\frac{1}{x^2}+\frac{1}{n}\frac{nx\sin(2nx)-2\sin^2(nx)}{x^3}\\=\frac{-nx+nx\sin(2nx)+\cos(2nx)-1}{nx^3}\\=\frac{nx(\sin(2nx)-1)+(\cos(2nx)-1)}{nx^3}\\<0\,\,$for$\,\,x>0\\$since$\,\,-1\leq \sin(2nx)\leq 1\,\,$and$\,\,-1\leq \cos(2nx)\leq 1\\$and when$\,\,\sin(2nx)=1,\cos(2nx)\neq 1[/maths]
trebla, put us out of our misery.Would've been much simpler if you just do:
Question:
Suppose f(x) = 1/x + sin²(πx) / πx². By considering the first derivative, prove that f(x) is decreasing for x > 0.
stfueveryone on this site is shit. they cannot solve my question
no ustfu
lol, u got meno u
for ur shitness in mathslol, u got me
no.for ur shitness in maths
?everyone on this site is shit. they cannot solve my question
Don't think it's pi; trebla is not in the habit of making such a typo.trebla, put us out of our misery.
is it a pi or an n?
thanks.
much agreed xDDon't think it's pi; trebla is not in the habit of making such a typo.
Provided n > 0 it's all ok. So azureus88's solution stands.
its not a typo, drongoski.Don't think it's pi; trebla is not in the habit of making such a typo.
Provided n > 0 it's all ok. So azureus88's solution stands.