Question

How do you find all points on the graph of `f(x)=sin^2x` at which the tangent line is horizontal?

Answer 1

graph{(sinx)^2 [-10, 10, -5, 5]}

`kpi/2 \ \ \ ;\ \ \ kinZZ`

Explanation:

`f^'(x)=2sinxcosx=sin2x`

`sin(2x)=0\ \ \` when `\ \ \ \2x=kpi\ \ \ ; \ \ \ k in ZZ`