Question

A function satisfying the conditions of Lagrange's mean value theorem on the interval $[1, 5]$ is:

• A. $f(x) = |x - 1| + |x - 5|$
• B. $g(x) = x - [x]$ ($[x]$ represents integral part of $x$)
• C. $h(x) = \begin{cases} 3x - 1, & 0 \le x \le 1 \\ x^2 + 1, & 1 \le x \le 4 \\ 17, & 4 \le x \le 6 \end{cases}$
• D. $k(x) = \begin{cases} x^2, & -\infty < x \le 1
  x, & 1 \le x \le \infty \end{cases}$

Hint:

Always remember that LMVT requires differentiability on the open interval $(a, b)$. Modulus functions are non-differentiable only where their inner expression becomes zero. For $|x-1| + |x-5|$, the non-differentiable points are $x=1$ and $x=5$, which lie exactly on the boundaries, keeping the open interval $(1, 5)$ perfectly differentiable!

Answer 1

The Correct Option is A

Concept: Lagrange's Mean Value Theorem (LMVT) states that if a function $\phi(x)$ is: 1) Continuous on the closed interval $[a, b]$, and 2) Differentiable on the open interval $(a, b)$, then there exists at least one $c \in (a, b)$ such that $\phi'(c) = \frac{\phi(b) - \phi(a)}{b - a}$. We must analyze each given option to see which function meets both conditions on the specific interval $[1, 5]$.

Step 1: Analyze Option (A): $f(x) = |x - 1| + |x - 5|$.
The absolute value functions $|x-1|$ and $|x-5|$ are continuous everywhere on $\mathbb{R}$, so their sum $f(x)$ is continuous on the closed interval $[1, 5]$. Now let us check differentiability on the open interval $(1, 5)$. Inside $(1, 5)$, we have $x > 1$ and $x < 5$. Therefore, $x - 1 > 0 \implies |x - 1| = x - 1$, and $x - 5 < 0 \implies |x - 5| = -(x - 5) = 5 - x$. Thus, for any $x \in (1, 5)$: \[ f(x) = (x - 1) + (5 - x) = 4 \] Since $f(x)$ reduces to a constant function $f(x) = 4$ on the open interval $(1, 5)$, its derivative is $f'(x) = 0$, which exists everywhere inside $(1, 5)$. Thus, $f(x)$ is continuous on $[1, 5]$ and differentiable on $(1, 5)$. LMVT conditions are satisfied.

Step 2: Analyze Option (B): $g(x) = x - [x]$.
The function $g(x) = x - [x]$ is the fractional part function $\{x\}$. This function is discontinuous at all integer points. Since the interval $[1, 5]$ contains the integers $2, 3, 4, 5$, $g(x)$ is not continuous on $[1, 5]$. Thus, LMVT fails.

Step 3: Analyze Option (C): $h(x)$.
Let us check continuity of $h(x)$ at the boundary points $x = 1$ and $x = 4$: - At $x = 1$: $\lim_{x \to 1^-} h(x) = 3(1) - 1 = 2$, and $\lim_{x \to 1^+} h(x) = 1^2 + 1 = 2$. It is continuous at $x = 1$. - At $x = 4$: $\lim_{x \to 4^-} h(x) = 4^2 + 1 = 17$, and $\lim_{x \to 4^+} h(x) = 17$. It is continuous at $x = 4$. Now check differentiability on $(1, 5)$, specifically at the internal point $x = 4$: - Left-hand derivative at $x = 4$: $\frac{d}{dx}(x^2 + 1) = 2x \implies 2(4) = 8$. - Right-hand derivative at $x = 4$: $\frac{d}{dx}(17) = 0$. Since $8 \neq 0$, $h(x)$ is not differentiable at $x = 4 \in (1, 5)$. Thus, LMVT fails.

Step 4: Analyze Option (D): $k(x)$.
Let us check differentiability at the internal point $x = 1 \in (1, 5)$: - Left-hand derivative at $x = 1$: $\frac{d}{dx}(x^2) = 2x \implies 2(1) = 2$. - Right-hand derivative at $x = 1$: $\frac{d}{dx}(x) = 1$. Since $2 \neq 1$, $k(x)$ is not differentiable at $x = 1$. Since $x = 1$ is the boundary point of our interval $[1, 5]$, differentiability on the open interval $(1, 5)$ holds true because inside $(1, 5)$, $k(x) = x$, which is differentiable. However, let us re-examine the continuity at $x = 1$: $\lim_{x \to 1^-} k(x) = 1^2 = 1$ and $\lim_{x \to 1^+} k(x) = 1$. So it is continuous at $x=1$.