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aspifsGak5u

Answered question

2021-12-11

Find the absolute maximum and minimum values of f on the given closed interval, and state where those values occur

f (x) = x - 2 \sin x; [- \frac{π}{4}, \frac{π}{2}]

SlabydouluS62

Skilled2021-12-12Added 52 answers

Step 1
Given function is

f (x) = x - 2 \sin x, [- \frac{π}{4}, \frac{π}{2}]

Step 2
Find the derivative of the function

f^{'} (x) = \frac{d}{d x} (x - 2 \sin x)

f^{'} (x) = 1 - 2 \cos x

For critical points
f'(x)=0

1 - 2 \cos x = 0

\cos x = \frac{1}{2} = \cos \frac{π}{3}

\Rightarrow x = \frac{π}{3} \in [- \frac{π}{4}, \frac{π}{2}]

Step 3
Now for maxima minima check the sign of f''(x) at that critical point

f^{″} (x) = \frac{d}{d x} (1 - 2 \cos x)

f^{″} (x) = 2 \sin x

x = \frac{π}{3} \Rightarrow f^{″} (\frac{π}{3}) = 2 \sin \frac{π}{3} = \sqrt{3} > 0

So,

x = \frac{π}{3}

is point of minima.
Step 4
So, the absolute minimum value is

f (\frac{π}{3}) = \frac{π}{3} - 2 \sin \frac{π}{3}

f (\frac{π}{3}) = \frac{π}{3} - 2 (\frac{\sqrt{3}}{2})

f (\frac{3}{2}) = \frac{π}{3} - \sqrt{3}

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