Find the maximum and minimum value of the function y-sin ^ 2 (x) -x / 2 on the interval [0, pi / 2]
July 29, 2021 | education|
Let’s find the derivative of the function:
y ‘= (sin ^ 2 (x) – x / 2)’ = 2sin (x) cos (x) – 1/2.
We equate it to zero:
2sin (x) cos (x) – 1/2 = 0.
Using the double argument formula for sine we get:
sin (2x) – 1/2 = 0;
sin (2x) = 1/2.
The roots of an equation of the form sin (x) = a are determined by the formula: x = arcsin (a) + – 2 * π * n, where n is a natural number.
2x = arcsin (1/2) + – 2 * π * n;
x = π / 12 + – π * n.
One point belongs to a given interval:
x0 = π / 12.
The value of the function at this point will be:
f (π / 12) = (sin (π / 12)) ^ 2 – π / 24.
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