Question

Let \( f \) be any continuous function on \( [0, 2] \) and twice differentiable on \( (0, 2) \). If \( f(0) = 0, f(1) = 1 \) and \( f(2) = 2 \), then :

• A. \( f'(x) = 0 \) for some \( x \in [0, 2] \)
• B. \( f''(x)>0 \) for all \( x \in (0, 2) \)
• C. \( f''(x) = 0 \) for some \( x \in (0, 2) \)
• D. \( f''(x) = 0 \) for all \( x \in (0, 2) \)

Hint:

If \( n+1 \) points are collinear for a function that is \( n \)-times differentiable, then the \( n \)-th derivative must vanish at some point in the interval containing those points.
Here, \( (0,0), (1,1), (2,2) \) are collinear, so \( f''(x) = 0 \) at some point.

Answer 1

The Correct Option is C

To solve this problem, we will utilize the Rolle's Theorem and its implications about the function's derivatives.

First, recall the conditions for Rolle's Theorem:

1. The function \( f \) must be continuous on the closed interval \([a, b]\). 
2. It must be differentiable on the open interval \((a, b)\).
3. \(f(a) = f(b)\).

In this problem, we are given:

• \( f(0) = 0 \)
• \( f(1) = 1 \)
• \( f(2) = 2 \)

Clearly, \( f(a) \neq f(b) \) for any choice of points in \([0, 2]\). Therefore, Rolle's theorem does not apply directly in its standard form for any part of this interval.

 

However, what we can consider is the Mean Value Theorem (MVT), which is a generalization of Rolle's Theorem. MVT states that if a function is continuous on \([a, b]\) and differentiable on \((a, b)\), there exists a \( c \in (a, b) \) such that:

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

For the intervals \([0, 1]\), \([1, 2]\), and \([0, 2]\), let us compute the derivative using MVT:

• For \([0, 1]\):

\[ f'(c_1) = \frac{f(1) - f(0)}{1 - 0} = \frac{1 - 0}{1} = 1 \] for some \( c_1 \in (0, 1) \).

• For \([1, 2]\):

\[ f'(c_2) = \frac{f(2) - f(1)}{2 - 1} = \frac{2 - 1}{1} = 1 \] for some \( c_2 \in (1, 2) \).

Since the average rate of change in derivative is constant in these intervals, but we have an additional information component given by continuity and the conditions of twice differentiability, by invoking the Mean Value Theorem for derivatives, we can infer that between every two values of \( f'(x) \) that are equal, there must be some value \( x \in (0, 2) \) such that:

• It implies that the second derivative \( f''(x) \) must hit zero at some point, since for continuous increase maintaining the constant speed, the acceleration momentarily becomes 0 at some point.

Thus, the correct option is: \( f''(x) = 0 \) for some \( x \in (0, 2) \).