Using Mean Value theorm for approximating points

Question

f(x) = sin^-1(x); [0,0.5] (help please!)?

Answer 1

Given function f(x) = sin^-1x

Given interval [ 0 , 0.5 ]

a = 0 , b = 0.5

f'(x) =1/√(1-x²)

By using mean value theorem,if f(x) is continuous on the interval [ 0 , 0.5 ] ,

f(x) is differentiable on ( 0 , 0.5 ) , then there is at least one number c

in the interval ( 0 , 0.5 ) ( that is 0 < c < 0.5 ) such that

 

f(a) = f(0) = sin^-10 = 0

f(b) = f(0.5) = sin^-10.5 = π/6

f'(c) = [ f(b) - f(a) ] / ( b - a )

f'(c) = [ f(0.5) - f(0) ] / ( 0.5- 0 )

f'(c) = [ π/6 - 0 ] / (0.5 - 0)

f'(c) = [ π/6] / (1/2)

1/√(1-c²) = 2π/6

1/√(1-c²) = π/3

√(1-c²) = 3/π

1-c²) = 9/π²

c² = 1 - 9/π²

c = √ (1 - 9/π²)

c = √ (1 -0.91)

c = 0.3