Question

f'(x)=sqrt(x)*sin(x)

The first derivative of the function f is given above. If f(0)=0, at which value of x does the function f attain it's minimum value on the closed interval [0,10]?

I know the answer is 6.28, but I need steps as to why. Please and thank you.

Answer 1

At the max/min of f(x), f ' (x) = 0 , so

?x sinx = 0
?x=0 ---> x = 0
or
sinx = 0 ---> x = 0, ?, 2?

We have to find f(x)
? ?x sinx dx ??? ---> nasty

tried integration by parts, got messy

how about this approach?
?xsinx = ? [?x(x - x^3/3! + x^4/4! - x^5/5! + ...]
using MacLaurin series
= ? [x^(3/2) - x^(7/2)/6 + x^(11/2)/24 + ...+ a constant ]
argghhh!! , have to integrate that.

ok, Wolfram to the rescue:
http://www.wolframalpha.com/input/?i=integral+%E2%88%9Ax+sinx

look at the graph of the integral, which they sketched to 2?
notice that 2? = appr 6.28
and the graph appears to have a minimum when x = 2?

Hopefully one of our other tutors will look at this and can see how to obtain the integral

Answer 2

forget the integral. It's not elementary.

besides, no one asked what f(x) was. The question was, where does it achieve its minimum value?