Question

Find the relative extrema using both first and second

derivative tests. f(x) = sin 2x, 0 < x < π

Answer 1

f'(x)=2cos2x=0

$2x=π/2+πn, n=0,1,2...$

$x=π/4+πn/2, n=0,1,2...$

f''(x)=-4sin2x

f''(π/4+πn/2)=-4sin(π/4+πn/2)=-2.84<0

So x=π/4+πn/2,n=0,1,2...x=π/4+πn/2,n=0,1,2 is minimum