Question

Hi folks, calculus question approaching.

\[f(x) = 8x+sin(4x)+4sin(2x)\]
So I go about finding the first derivative for part 1 of the question:
\[f'(x)=8+4cos(4x)+8cos(2x)\], which was found in terms of cos(2x) for part 2 of the question (not necessarily correct):
\[f'(x)=8cos^{2}(2x)+8cos(2x)+4\]
Now part 3 is to prove that the function f(x) is always increasing. So I figure this is true if f'(x) is always positive, i.e. \[8cos^{2}(2x)+8cos(2x)+4 >0\]As far as I can see, this cannot be factorised, and if it were an equation, it would have complex roots.

Answer 1

The best I've managed is that \[cos^{2}(2x)+cos(2x) > -\frac{1}{2}\], or \[cos(2x)[cos(2x)+1] > -\frac{1}{2}\], which I'm not convinced are true statements.

Answer 2

Try graphing the original function, this graph appears to be always increasing.

Answer 3

I think he needs a rigorous proof, Chaise.

Answer 4

Well. Your derievative is the form of an quadratic whose discriminant is less than zero.
=> That it is always above or always below the x axis ( No roots)
Since The coefficient of cos^2 2x is greater than zero => the entire graph is above x axis.
Therefore it is always increasing.

Answer 5

Yeah, I threw it into wolfram alpha, and it is always increasing as far as I can tell.

Answer 6

what @siddhantsharan , said is correct... your derivative is written in quadratic form... the discriminant is negative...

Answer 7

Sid, nice proof. I like that you took principles from polynomials. :)

Answer 8

Thanks Sid, looks like that will work fine. I was rather hoping to work algebraically through the inequality to come to a definite statement about the slope, e.g. cos(2x)>-2 or something like that. Oh well! Thanks for the help.

Answer 9

Haha. Thanks!