Question

Verify Lagrange's mean value theorem for the function \( f(x) = \sqrt{x + 4} \) on the interval \([0, 5]\).

Hint:

Lagrange's Mean Value Theorem guarantees the existence of a point where the instantaneous rate of change equals the average rate of change over the interval.

Answer 1

Step 1: Lagrange's Mean Value Theorem states that if \( f(x) \) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists at least one \( c \in (a, b) \) such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \] Step 2: For the given function \( f(x) = \sqrt{x + 4} \), it is continuous and differentiable on \([0, 5]\) because it is a polynomial function. Step 3: The interval is \([0, 5]\), so we compute \( f(0) \) and \( f(5) \): \[ f(0) = \sqrt{0 + 4} = 2, \quad f(5) = \sqrt{5 + 4} = 3 \] Step 4: Using Lagrange’s mean value theorem: \[ f'(c) = \frac{f(5) - f(0)}{5 - 0} = \frac{3 - 2}{5} = \frac{1}{5} \] Step 5: Now, compute \( f'(x) \). The derivative of \( f(x) = \sqrt{x + 4} \) is: \[ f'(x) = \frac{d}{dx} \left( \sqrt{x + 4} \right) = \frac{1}{2\sqrt{x + 4}} \] Step 6: Set \( f'(c) = \frac{1}{5} \) and solve for \( c \): \[ \frac{1}{2\sqrt{c + 4}} = \frac{1}{5} \] \[ 2\sqrt{c + 4} = 5 \] \[ \sqrt{c + 4} = \frac{5}{2} \] \[ c + 4 = \left( \frac{5}{2} \right)^2 = \frac{25}{4} \] \[ c = \frac{25}{4} - 4 = \frac{25}{4} - \frac{16}{4} = \frac{9}{4} \] Step 7: Thus, \( c = \frac{9}{4} \), and we have verified that the Lagrange Mean Value Theorem holds for the function on the interval \([0, 5]\).