I have the following function:
$$ f(x) =\sin(x) – \sqrt{3} * \cos(x) $$ and $$ I = [-\pi , +\pi] $$
I should check this function for monotonicity and convexity. So I drawed the $\sin$ and $\cos$ on a paper to see how the curve goes between $-\pi$ and $+\pi$. but the whole function is a sum between $\sin$ and $\cos$, so how do I do that? My brain stops there. I surely can see where the sin or cos for itself is monotonic but don't know how to add them together, and the $\sqrt3$ gets me confused too.
$$ f'(x) = \cos(x) + \sqrt3 * \cos(x) \\
f''(x) = -\sin(x) +\sqrt3 * \cos(x) $$
I then made first and 2nd derivative. But how will i go further? I know where the sin(x) is monotonic .. but I don't know where $\sin x + \cos x$ is. I'm really having a bad time combining and imagining these 2 together..
Or is there just a 100% mathematical approach without looking at the graphs?
Best Answer
It seems the following.
This is the graph of the function $f$ and its derivatives. So intervals of monotonicity and convexity of the function $f$ are those of constant signs of the first derivative and the second, respectively. To find them we should to solve the equalities $fâ(x)=0$ and $fââ(x)=0$. For this purpose we use the standard approach.
$$a\sin x+b\cos x=\sqrt{a^2+b^2}\sin (x+y)= \sqrt{a^2+b^2}(\sin x\cos y+\cos x\sin y).$$
From here we have $\cos y=a/\sqrt{a^2+b^2}$ and $\sin y=b/\sqrt{a^2+b^2}$. In our particular case, $\cos y=1/2$, $\sin y=-\sqrt 3/2$, so we may put $y=-\pi/3$. So we have $f(x)=2\sin (x-\pi/3)$.