The Mean Value Theorem states:
(i) For some function f(x) which is continuous and differentiable on an interval [a,b], some c exists such that the slope of the line connecting (a,f(a)) and (b,f(b)) is equal to the slope of the line tangent to f(x) at x=c.
First, let's note that in our case, f(x) is equal to sqrt(1-x^2), and the interval [a,b] that we're looking at is [0,1]. Let's now find the slope of the secant line connecting (0, f(0)) to (1, f(1)).
m = (f(1) - f(0))/(1 - 0)
We know f(1) is 0, and f(0) is 1, so
m = (0 - 1)/(1 - 0) = -1
So, the slope of the secant line, or in other words, the average rate of change, on [0, 1] is equal to -1.
The Mean Value Theorem, in this case, assures us that some c exists such that -1 is equal to the slope of the line tangent to f(x) at x=c. Or in other words, -1 is equal to f'(c).
All we need to do, to find the value c that satisfies the Mean Value Theorem, is differentiate f(x) and set it equal to -1:
-1 = d/dx[sqrt(1 - x^2)]
This is a simple application of the chain rule:
-1 = 1/2*(1-c^2)^(-1/2)*(-2c)
Simplification gives us:
1 = c/sqrt(1-c^2)
To solve for c, we will simply multiply both sides by sqrt(1-c^2) and then square both sides. This yields:
1 - c^2 = c^2
Now we can easily solve for c:
c = sqrt(2)/2
So what we've shown is that when x = c = sqrt(2)/2, the instantaneous rate of change of f(x) equals -1, which is also the average rate of change across [0, 1]. The Mean Value Theorem assured us that some c exists that satisfies this condition, and we've gone and found exactly the c that satisfies it.