A Mean Value Theorem or MVT Question?

Question

Find all numbers c between a and b for which:
(f(b)-f(a))/(b-a) = f'(c)

The Problem:
f(x) = (x)arcsin(x) on [0,1]

Answer

Let’s start off with a definition.

The Mean Value Theorem is one of the most important theoretical tools in Calculus. It states that if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that
f ‘(c) = ( f(b) – f(a) ) / ( b – a)

OK.
So in our problem here, a = 0 and b = 1. Thus,
f(a) = f(0) = 0
and f(b) = f(1) = arcsin(1) = (pi)/2

f ‘(c) = ( f(b) – f(a) ) / ( b – a)

= ( f(1) – f(0) ) / ( 1 – 0)

= ( (pi)/2 – 0 )

= (pi)/ 2

Now,
f'(x) = x [ 1 / (1 – x^2)^{1/2} ] + arcsin(x)

using the fact if g(x) = arcsin(x) Then,
g'(x) = 1 / (1 – x^2)^{1/2}
and the multiplication rule in differentiation.

Hence, f'(c) = c [ 1 / (1 – c^2)^{1/2} ] + arcsin(c)
and f'(c) = (pi)/2

Now, note another fact that
arcsin(x) = x / (1 – x^2)^{1/2}

Hence,

c [ 1 / (1 – c^2)^{1/2} ] + c / (1 – c^2)^{1/2} = (pi)/2

That is,

( 2c ) / ( 1 – c^2 )^{1/2} = (pi)/2

Manipulating using algebra, we get,

( 1 – c^2 ) = [ 16 c^2 ] / [(pi)]
Thus,
c^2 = [ (pi) ] / [ (pi) + 16 ]

For our answer for the value c, we only take the positive square root because, the question asked for c in the interval [0,1].

Hence, using a calculator, you should get…

c = 0.405122068…