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

Question

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

Answer 1

Min value?

f'(x)=0=sqrt(x)*sin(x)
well at x=0 is solution.
sin(x)=0
x= n*PI is a solution up to
n<10/PI for n=1,2,3 test now second derivative...
f"(x)=cos(x)sqrtx + 1/2sqrtx *sinx
at x=0, is zero
at x= PI, is negative
at x=2PI, is positive
at x=3 pi, is negative

Minimum then by definition is at x=2PI, not counting the initial value at x=0
check my thinking.

Answer 2

looks good. f(x) has a min when f' changes from - to + (f bottoms out and starts back up)

That happens at x=2pi