Global extreme question. Please help! :)

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

The first derivative is 1 - sqrt(2)cos(x). Equate to 0 and solve for x. You get cos(x) = 1/sqrt(2). The only solution in [0, Pi] is Pi/4. Plug this x and the end points of the given intervals in the given function. The outputs are 0, Pi/4 - 1, Pi. The min and max values are Pi/4 - 1 and Pi. These are the global extremes.

Answer 2

First find the derivative of the function:

Φ ' (x) = 1-√2 * cos x

The set derivative = 0 and solve:

1-√2 * cos x=0

1=√2 * cos x

1/√2 = cos x

x = π/4 (since we are only looking from 0 to π)

Now we plug in that value and our two endpoint values into the original function to find the global maximum and minimum:

Φ (0) = 0-√2 * sin (0) = 0

Φ (π/4) = π/4-√2 * sin (π/4) = π/4 - √2 * 1 = π/4 - √2 = approximately -0.6

Φ (π) = π-√2 * sin (π) = π - √2 * 0 = π

The global max of our function is π. The global min of our function is π/4 - √2