Maximum value of (arcsinx)^2 - 2arcsinx

f(x) has a global maximum at x = -1

f(x) has a global minimum at x = sin(1)

Step-by-step 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:

f(x) has a global maximum at x = -1

f(x) has a global minimum at x = sin(1)

10 cm

step-by-step explanation:

if the polygon is a rectangle, then the perimeter is x+x+y+y. using the values that you gave, the perimeter is 3+3+2+2=10cm.

5y

step-by-step explanation: