Answer:
f(x) has a global maximum at x = -1
f(x) has a global minimum at x = sin(1)
Explanation:
Find and classify the global extrema of the following function:
f(x) = sin^(-1)(x)^2 - 2 sin^(-1)(x)
Find the critical points of f(x):
Compute the critical points of sin^(-1)(x)^2 - 2 sin^(-1)(x)
The domain of f(x) = sin^(-1)(x)^2 - 2 sin^(-1)(x) is {x element R : -1<=x<=1}
The interior of the domain is {x element R : -1<x<1}:
f(x) = sin^(-1)(x)^2 - 2 sin^(-1)(x) when -1<x<1
To find all critical points, first compute f'(x):
d/( dx)(sin^(-1)(x)^2 - 2 sin^(-1)(x)) = (2 sin^(-1)(x))/sqrt(1 - x^2) - 2/sqrt(1 - x^2):
f'(x) = (2 sin^(-1)(x))/sqrt(1 - x^2) - 2/sqrt(1 - x^2)
Solving (2 sin^(-1)(x))/sqrt(1 - x^2) - 2/sqrt(1 - x^2) = 0 yields x = sin(1):
x = sin(1)
f'(x) exists for all x such that -1<x<1:
(2 sin^(-1)(x))/sqrt(1 - x^2) - 2/sqrt(1 - x^2) exists for all x such that -1<x<1
The only critical point of sin^(-1)(x)^2 - 2 sin^(-1)(x) is at x = sin(1):
x = sin(1)
The domain of sin^(-1)(x)^2 - 2 sin^(-1)(x) is {x element R : -1<=x<=1}:
The endpoints of {x element R : -1<=x<=1} are x = -1 and 1
Evaluate sin^(-1)(x)^2 - 2 sin^(-1)(x) at x = -1, sin(1) and 1:
x | f(x)
-1 | 5.60899
sin(1) | -1
1 | -0.674192
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:
x | f(x) | extrema type
-1 | 5.60899 | global max
sin(1) | -1 | global min
1 | -0.674192 | neither
f(x) = sin^(-1)(x)^2 - 2 sin^(-1)(x) has one global minimum and one global maximum:
Answer: f(x) has a global maximum at x = -1
f(x) has a global minimum at x = sin(1)