# Locating the absolute extreme of the function on the closed interval

Locate the absolute extreme of the function on the closed interval

$$f(x) = \sin (\pi x)$$ on $\displaystyle[\frac{-7}{6}, \frac{-1}{2}]$.

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This is a classic style problem. I suspect that you have an idea of what to do. What have you tried and where did you get stuck? –  mixedmath Jan 6 '13 at 20:58

Note that between $[-7/2,-1/2]$ the function $\cos(\pi x)\leq 0$. This means that the continuous differentiable function $f(x)$ is decreasing on this closed interval. So it takes its absolute extreme at the end points of the interval. Be careful of relative extreme. They are as @Daniel noted above when $k=\pi$. Your interval has no such these kind of extreme.
Hint: Since $f$ is continuous and the interval is compact, we can be sure that the maximum and minimum are inded assumed. The extrema are either at boundary points of the interval or in its interior. In the latter case, standard methods for finding local extrema for differentiable functions apply.
While there are standard tools for differentiable functions in general, it is probably worth noting that the extrema for $\sin(kx)$ are known to be at $(n+\frac{1}{2})\frac{\pi}{k}$ for integer $n$ –  Daniel Littlewood Jan 6 '13 at 21:04