Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Locate the absolute extreme of the function on the closed interval

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

share|improve this question
2  
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
add comment

2 Answers 2

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.

share|improve this answer
add comment

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.

share|improve this answer
    
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
    
Am I allowed to add a picture here in your answer for the OP? You can edit and remove my name, Hagen. :) –  B. S. Jan 6 '13 at 21:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.