Please help!! Very hard calc question.

Question

Let φ: [0,π] → ℝ; x ↦ φ(x) = x –√(2)sin(x). Find the global extreme values of φ on [0,π].

(The only thing under the sqr rt is 2)

Answer 1

Hi Ruby,

All you need to do is take the derivative, which is easy since the derivative of x is 1 and the derivative of sin(x) is cos(x). Then set the derivative equal to 0 to find any critical points in the interval. Then just evaluate the function at these critical points and the endpoints of the interval. The largest of these values is the maximum value on the interval, and the smallest is the minimum.