f(x) = integral from 1/3 to x of cos (1/t squared) dt for 1/3 < or equal to x < or equal to 1.

Question

what values of x does f attain a relative maximum?

Answer 1

f(x)=_⅓∫^xcos(1/t²)dt, df/dx=cos(1/x²)=0 when f(x) is max or min. We don't need to integrate the t function because we need to differentiate an integration which means we can use the integrand to test for extrema. The integration generates a function of x plus a constant term, which disappears on differentiation.

Cosine is 0 when its argument is (2n+1)π/2 where n is any integer. The value of cos(1/x²) when x=⅓ is cos(9)=-0.9111 (approx) where 9 is radians. This means that the gradient of f is negative when x is at the lower limit of the interval.

1/x²=π/2 when cos(1/x²)=0. Therefore x=√(2/π)=0.7979 approx. This is the value of x when f has a max or min. We need to differentiate again to find the nature of this extremum.

d²f/dx²=-sin(1/x²)(-2/x³)=2sin(1/x²)/x³=2/x³>0 so x=0.7979 is a minimum for f. This value of x is in the given range.

The next extremum is when 1/x²=3π/2, so x=√(2/3π)=0.4607 approx. This value of x is also in range.

However, d²f/dx²<0 for this value of x because sin(3π/2)=-1. Therefore f has a maximum at x=√(2/3π).

The next extremum is when x=√(2/5π)=0.3568 approx, still in range but this is a minimum for f.

The next extremum is when x=√(2/7π)<⅓, so it is out of range (this would in fact be a maximum). Since the gradient of f is negative when x=⅓, f would be decreasing from maximum at x=√(2/7π) to its minimum at x=√(2/5π).

Even though we don't know (or haven't found) f(x), we do know its x values for its max and min between x=⅓ and x=1: x=√(2/5π) (min), √(2/3π) (max), √(2/π) (min).