Question

Which of the following function satisfies hypotheses and the conclusion of the Lagrange Mean Value Theorem

• A. $f(x) = |x|$ on $[-1, 1]$
• B. $f(x) = \sqrt{x}$ on $[-1, 1]$
• C. $f(x) = \sqrt[3]{x}$ on $[-1, 1]$
• D. $f(x) = 2x^2 - 7x + 10$ on $[2, 5]$

Hint:

Polynomials, exponential functions, $\sin(x)$, and $\cos(x)$ are everywhere continuous and differentiable. They always satisfy LMVT on any finite interval $[a,b]$. Watch out for $|x|$, $x^{1/3}$, or fractions with zero in denominators inside $(a,b)$.

Answer 1

The Correct Option is D

Step 1: Concept
Lagrange's Mean Value Theorem (LMVT): Let $f: [a, b] \to \mathbb{R}$ be a function. LMVT requires two hypotheses:
1. $f(x)$ is continuous on the closed interval $[a, b]$.
2. $f(x)$ is differentiable on the open interval $(a, b)$.
Under these conditions, there exists at least one $c \in (a, b)$ such that: \[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

Step 2: Key Formulas and Approach
We check continuity and differentiability for each function on its given interval. Polynomials are infinitely differentiable everywhere on $\mathbb{R}$, making them automatic candidates.

Step 3: Step-by-step Explanation

• Option (A): $f(x) = |x|$ on $[-1, 1]$
The absolute value function $f(x) = |x|$ is continuous on $[-1, 1]$, but it is not differentiable at $x = 0 \in (-1, 1)$. Hence, LMVT hypothesis fails.

• Option (B): $f(x) = \sqrt{x$ on $[-1, 1]$}
$f(x) = \sqrt{x}$ is undefined for $x < 0$ in the real number system. Thus it is not even defined on $[-1, 0)$. LMVT hypothesis fails.

• Option (C): $f(x) = \sqrt[3]{x = x^{1/3}$ on $[-1, 1]$}
The derivative is $f'(x) = \frac{1}{3 x^{2/3}}$. At $x = 0 \in (-1, 1)$, $f'(0)$ does not exist (vertical tangent). Thus $f(x)$ is not differentiable on $(-1, 1)$. LMVT hypothesis fails.

• Option (D): $f(x) = 2x^2 - 7x + 10$ on $[2, 5]$
Since $f(x)$ is a polynomial function:
1. It is continuous on $[2, 5]$.
2. It is differentiable on $(2, 5)$ with $f'(x) = 4x - 7$.
LMVT hypotheses are fully satisfied!
Conclusion verification:
\[ \frac{f(5) - f(2)}{5 - 2} = \frac{(2(25) - 35 + 10) - (2(4) - 14 + 10)}{3} = \frac{25 - 4}{3} = \frac{21}{3} = 7 \] Set $f'(c) = 7 \implies 4c - 7 = 7 \implies 4c = 14 \implies c = 3.5 \in (2, 5)$. Both hypotheses and conclusion hold perfectly.

Step 4: Final Answer
The polynomial function $f(x) = 2x^2 - 7x + 10$ on $[2, 5]$ satisfies all conditions of LMVT. Thus, Option (D) is correct.