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.
