f'(x) = (x)^(1/2)sin x
f(0) = 0
Minimum value of f(x) in [0, 10]?

Question

f'(x) = (x)^(1/2)sin x
f(0) = 0
Minimum value of f(x) in [0, 10]?

anonymous · Answer

Look at he graph to help you decide about the Minimum fo f on [0,10] http://www.wolframalpha.com/input/?i=plot++sqrt%28x%29+sin%28x%29+frm+x%3D0+to+x%3D10

anonymous · Answer

What do you think the answer is? 

This question is asked in the CLEP Calculus Study Guide, and I don't think the graph even comes close to the supposed answer...

anonymous · Answer

It does not seem feasible by setting the derivative equal to zero. I found it numerically. The minimum occurs at x=4.81584 and  the minimum is -2.18277

anonymous · Answer

Is your function
$$
\sqrt x \, \sin(x)
$$

anonymous · Answer

Yes. The answer is 6.28 btw.

anonymous · Answer

Something to do with 2pi probably?

anonymous · Answer

It is not 6.2 for the function $\Large \sqrt x \, \sin(x) $ on [0,10]

anonymous · Answer

The graph tells the whole story.

anonymous · Answer

Hmm.. fair enough.

anonymous · Answer

Are you sure you copied the function correctly?

anonymous · Answer

Yes.. it is $$\sqrt{x}\sin x$$

anonymous · Answer

How can you evaluate this without using a graph?

anonymous · Answer

You cannot evaluate is explicitly but numerically http://www.wolframalpha.com/input/?i=find+minimum++sqrt%28x%29+sin%28x%29+frm+x%3D0+to+x%3D10

anonymous · Answer

Sorry, I thought $f(x)= \sqrt x \sin(x)    $ , I did not see that f'(x) was equal that.
Hold on a minute

anonymous · Answer

Yup, that's the case. No worries.

anonymous · Answer

If you look at the garph of  f'(x), you can see that before x=6,28, that f'(x) is negative and after that f'(x) is positive, which means that f has a minimum at x=6.28

anonymous · Answer

Here is a complete solution. To study the sign of f'(a) on [0, 10] it is enough to study the sign of sin(x) since $\sqrt x$ is always  positive on [0,10]
x                   0                  Pi                  2 pi             10
sin(x)             0         +      0          -        0         +
f'(x)                0         +      0          -        0         +
f(x)                          inc      Max      dec    Min     inc
You can see that the minimum occurs at x = 2pi

anonymous · Answer

Do you understand it now?

anonymous · Answer

Oh fantastic. Thanks a bunch.

anonymous · Answer

Okay, so basically.. if a function is decreasing between a certain range, and if one of the limits is the global maxima, the other has to be the minima?