# How do you find all numbers c that satisfy the conclusion of the mean value theorem for f(x) = sqrt(1-x^2) on the interval [0-1]?

Sep 11, 2015

$c = \frac{\sqrt{2}}{2}$

#### Explanation:

The Mean Value Theorem states:

(i) For some function $f \left(x\right)$ which is continuous and differentiable on an interval $\left[a , b\right]$, some $c$ exists such that the slope of the line connecting $\left(a , f \left(a\right)\right)$ and $\left(b , f \left(b\right)\right)$ is equal to the slope of the line tangent to $f \left(x\right)$ at $x = c$.

First, let's note that in our case, $f \left(x\right)$ is equal to $\sqrt{1 - {x}^{2}}$, and the interval $\left[a , b\right]$ that we're looking at is $\left[0 , 1\right]$. Let's now find the slope of the secant line connecting $\left(0 , f \left(0\right)\right)$ to $\left(1 , f \left(1\right)\right)$.

$m = \frac{f \left(1\right) - f \left(0\right)}{1 - 0}$

We know $f \left(1\right)$ is $0$, and $f \left(0\right)$ is $1$, so

$m = \frac{0 - 1}{1 - 0} = - 1$

So, the slope of the secant line, or in other words, the average rate of change, on $\left[0 , 1\right]$ 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 $f \left(x\right)$ at $x = c$. Or in other words, $- 1$ is equal to $f ' \left(c\right)$.

All we need to do, to find the value $c$ that satisfies the Mean Value Theorem, is differentiate $f \left(x\right)$ and set it equal to $- 1$:

$- 1 = \frac{d}{\mathrm{dx}} \left[\sqrt{1 - {x}^{2}}\right]$

This is a simple application of the chain rule:

$- 1 = \frac{1}{2} \cdot {\left(1 - {c}^{2}\right)}^{- \frac{1}{2}} \cdot \left(- 2 c\right)$

Simplification gives us:

$1 = \frac{c}{\sqrt{1 - {c}^{2}}}$

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 $c$:

$c = \frac{\sqrt{2}}{2}$

So what we've shown is that when $x = c = \frac{\sqrt{2}}{2}$, the instantaneous rate of change of $f \left(x\right)$ equals $- 1$, which is also the average rate of change across $\left[0 , 1\right]$. 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.