Finding the Value 'c' in the Mean Value Theorem: Solved Examples

📚 Quick Study Guide

🔍 The Mean Value Theorem (MVT) states that if a function $f(x)$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$, then there exists a point $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.

💡 To find the value 'c', follow these steps:
• Calculate $f'(x)$, the derivative of $f(x)$.
• Calculate $\frac{f(b) - f(a)}{b - a}$, the average rate of change of $f(x)$ over $[a, b]$.
• Set $f'(c) = \frac{f(b) - f(a)}{b - a}$ and solve for $c$.
• Ensure that the value of $c$ you find lies within the interval $(a, b)$.

Practice Quiz

1. What is the first step in finding the value 'c' guaranteed by the Mean Value Theorem for a function $f(x)$ on the interval $[a, b]$?
   1. Calculate $f(b) - f(a)$.
   2. Find $f'(x)$.
   3. Evaluate $f(a)$.
   4. Set $f'(c) = 0$.
2. Given $f(x) = x^2$ on the interval $[1, 3]$, what is the value of $\frac{f(b) - f(a)}{b - a}$?
   1. 2
   2. 3
   3. 4
   4. 5
3. If $f(x) = x^3$ on the interval $[0, 2]$, what is $f'(x)$?
   1. $x^2$
   2. $2x$
   3. $3x^2$
   4. $3x$
4. For $f(x) = x^2$ on $[0, 2]$, what equation do you need to solve to find 'c' after finding $f'(x)$ and $\frac{f(b) - f(a)}{b - a}$?
   1. $2c = 4$
   2. $c^2 = 2$
   3. $c = 4$
   4. $2c = 2$
5. Consider $f(x) = \sqrt{x}$ on the interval $[1, 4]$. Find the value of $f'(x)$.
   1. $\frac{1}{2\sqrt{x}}$
   2. $\frac{1}{\sqrt{x}}$
   3. $2\sqrt{x}$
   4. $\sqrt{x}$
6. Which condition MUST be satisfied for the Mean Value Theorem to be applicable to a function $f(x)$ on $[a, b]$?
   1. $f(x)$ must be continuous on $(a, b)$ only.
   2. $f(x)$ must be differentiable on $[a, b]$.
   3. $f(x)$ must be continuous on $[a, b]$ and differentiable on $(a, b)$.
   4. $f(a) = f(b)$.
7. For $f(x) = x^2 + 2x - 1$ on the interval $[0, 1]$, what is the value of $c$ guaranteed by the Mean Value Theorem?
   1. 0
   2. 0.5
   3. 1
   4. -1