The absolute maximum of n(x) = 1 - 2 · cos 2x, whose first derivative is n(0) = 0, exists at x = 0.5π on (0, 2π]. The value of the absolute maximum is 3.
A function is differentiable in a given interval if and only if its derivative exists for every value of the interval. To determine the absolute maximum of n(x), we need to integrate the given function, apply first derivative and second derivative test and finally evaluate n(x).
First, we integrate n'(x) in terms of x:
n(x) = 2 ∫ sin 2x dx = 2 · (- 0.5 · cos 2x) + C = - cos 2x + C
- cos 0 + C = 0
C = 1
Then, n(x) = 1 - 2 · cos 2x.
Now we proceed to apply first and second derivative tests:
2 · sin 2x = 0
sin 2x = 0
2 · x = sin⁻¹ 0
2 · x = π ∨ 2π
x = 0.5π ∨ π
n''(x) = 4 · cos 2x
n''(π) = 4 · cos 2π
n''(π) = 4
There is an absolute minimum for x = π. [tex]\blacksquare[/tex]
n''(0.5π) = 4 · cos π
n''(0.5π) = - 4
There is an absolute maximum for x = 0.5π. [tex]\blacksquare[/tex]
Finally, we evaluate n(x) for x = 0.5π to determine the absolute maximum:
n(0.5π) = 1 - 2 · cos π
n(0.5π) = 3
The absolute maximum of n(x) = 1 - 2 · cos 2x, whose first derivative is n(0) = 0, exists at x = 0.5π on (0, 2π]. The value of the absolute maximum is 3. [tex]\blacksquare[/tex]
The statement of the question is poorly formatted and reports several mistakes, correct form is introduced below:
If n'(x) = 2 · sin 2x and n(0) = 0. What is the absolute maximum of n(x) on (0, 2π]?
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