Minimum value of $y=\sin( 2x) - x$, where $x\in [-\frac{\pi}2,\frac{\pi}2]$

Question

I tried applying the concept that at minima, derivative of $y$ with respect to $x$ should be zero, but realised that it fails as the domain is restricted. Rightly, upon plotting the graph, we can see that we cannot equate $\frac{\mathrm dy}{\mathrm dx}$ to be zero. Can you help me arrive at the solution in a way that doesn't require graph plotting?

Answer 1

The derivative test claims that local extrema in the interior of the domain are to be found among the zeros of the derivative; as noted above, in this case $\pm{\frac{\pi}{6}}$ are the zeros of the derivative and it is easy to see that the plus one is a local maximum and the minus one is a local minimum.

However, the boundary points need also to be checked as there the function may have an extremum without the derivative being zero, and in this case it is trivial to see that $-\frac{\pi}{2}$ is the global minimum attained at $\frac{\pi}{2}$ since $\sin({2x}) \ge 0, x \in [0,\frac{\pi}{2}]$, hence $\sin({2x})-x \ge -x \ge -\frac{\pi}{2}$ there while the function is odd and for negative $x$ obviously at least $-1$ (actually the local minimum for negative $x$ is at $-\frac{\pi}{6}$ and hence it is $-\frac{\sqrt{3}}{2}+\frac{\pi}{6}=-0.34...$

Answer 2

$$\frac{dy}{dx}=2\cos(2x)-1=0$$ has the solutions

$$x=\pm\frac\pi3$$ in the specified interval.

Now compare the values of the function at the stationary points and interval endpoints.

Answer 3

Why can't you equate its derivative to 0??

$$\frac{dy}{dx}=2\cos(2x)-1=0$$ gives:

$$cos(2x)= \frac12$$

Hence, we obtain $$x=\pm\frac\pi6$$ which are in the specified interval.

Note that these are the points of local maxima and minima. The max and min over the entire given domain will obviously be at $-\frac\pi2$ and $\frac\pi2$ respectively.

Answer 4

Choice of the domain is restricted to the domain interval you stated as bounded between two limits. But everything else remains same.

Derivative $ \dfrac{dy}{dx}=2\cos(2x)-1 $ vanishes at an infinite number of points when domain is not restricted to $ x= \pm \dfrac{\pi}{3}.$ When you selected a truncated interval $|x|<\dfrac{\pi}{2} $ the derivative underwent sign change only twice.