Show that sinx<x for x>0.
Here's my working so far:
Let f(x)=sinx-x
f'(x)=cosx-1 =0 for stat pts
cosx=1
x=0,360
x>0 so x=360 degrees
f''(x)=-sinx
f''(360)=0
so apparently there's a horizontal point of inflexion at x=360
but don't you have to show that it's the absolute max turning point?
Firstly, for the sake of calculus, never use degrees.
Secondly, \(f^{\prime\prime}(x) = 0\) doesn't always imply that \(x\) is a point of inflexion. The theorem says that if \(x\) is a point of inflexion, then \(f^{\prime\prime}(x) = 0\). But the converse (i.e. what you get when you flip that statement around) does not hold. An easy example is \( h(x) = x^4\) - we can prove that \( h^{\prime\prime}(0) = 0\), but \(x=0\) is certainly not a point of inflexion.
To test if it is actually a point of inflexion, you need to do test both sides for concavity changes. So since you've computed that \(h^{\prime\prime}(2\pi) = 0\), you could plug \(2\pi - 0.01\) and \(2\pi + 0.01\) into \(f(x)\), to determine if there is a concavity change. You'll see that there is not, so the stationary point you found will not be a horizontal point of inflexion.
Finally, as for the original question itself, hint - you've overcomplicated it. Verifying that \(f^{\prime}(x) \leq 0\) for all real \(x\) is the starting point. Recall that if \(f^\prime(x) \leq 0\) for all real \(x\), i.e. the curve is decreasing (albeit not strictly decreasing), then we have the conclusion
\[ \text{If }x_1 \geq x_2\\ \text{Then }f(x_1) \leq f(x_2) \]