To apply the Mean Value Theorem (MVT) to the function $f(x) = x^2 + 2x - 1$ on the interval $[0,1]$, follow these steps:

1. State the conditions for the MVT:

The Mean Value Theorem applies if:

• $f(x)$ is continuous on $[a, b]$.
• $f(x)$ is differentiable on $(a, b)$.

Here, $f(x) = x^2 + 2x - 1$ is a polynomial, so it is continuous and differentiable everywhere. Therefore, the conditions for the MVT are satisfied.

2. Apply the formula of the MVT:

The MVT guarantees the existence of a point $c \in (0,1)$ such that: $f'(c) = \frac{f(b) - f(a)}{b - a}$ For the interval $[0,1]$, let $a = 0$ and $b = 1$. Compute the required values:

1. Find $f(a)$ and $f(b)$: $f(0) = (0)^2 + 2(0) - 1 = -1$ $f(1) = (1)^2 + 2(1) - 1 = 2$

2. Calculate the average rate of change: $\frac{f(1) - f(0)}{1 - 0} = \frac{2 - (-1)}{1 - 0} = \frac{3}{1} = 3$

3. Differentiate $f(x)$:

The derivative of $f(x) = x^2 + 2x - 1$ is: $f'(x) = 2x + 2$

4. Solve for $c$:

Find $c \in (0,1)$ such that: $f'(c) = 3$ $2c + 2 = 3$ $2c = 1 \quad \implies \quad c = \frac{1}{2}$

5. Conclusion:

By the Mean Value Theorem, there exists $c = \frac{1}{2}$ in the interval $(0,1)$ such that: $f'(c) = \frac{f(1) - f(0)}{1 - 0}$

Would you like additional details or explanations on this?

Follow-up Questions:

1. What does the Mean Value Theorem guarantee geometrically on the interval $[0,1]$?
2. How does the derivative at $c$ relate to the slope of the secant line?
3. Can you verify the result by graphing $f(x)$ and the secant line?
4. Does the Mean Value Theorem apply to non-polynomial functions like $f(x) = \sin(x)$ on $[0, \pi]$?
5. What happens if $f(x)$ is not differentiable at a point within the interval?

Tip: For polynomial functions, the Mean Value Theorem often has an easy solution since derivatives are straightforward to compute!