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

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

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Answer 1 · 2015-09-11T04:46:12

$c = \frac{\sqrt{2}}{2}$ $c = sqrt(2)/2$

Explanation:

The Mean Value Theorem states:

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

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

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

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

$m = \frac{0 - 1}{1 - 0} = - 1$ $m = (0 - 1)/(1 - 0) = -1$

So, the slope of the secant line, or in other words, the average rate of change, on $[0, 1]$ $[0, 1]$ is equal to $- 1$ $-1$ .

The Mean Value Theorem, in this case, assures us that some $c$ $c$ exists such that $- 1$ $-1$ is equal to the slope of the line tangent to $f (x)$ $f(x)$ at $x = c$ $x=c$ . Or in other words, $- 1$ $-1$ 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.

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

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How do you find all numbers c that satisfy the conclusion of the mean value theorem for f(x) = sqrt(1-x^2)f(x)=√1−x2f(x) = sqrt(1-x^2) on the interval [0-1]?

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