Key Concepts and Formulas

• Monotonicity using the First Derivative Test: If $\phi'(x) > 0$ on an interval, then $\phi(x)$ is increasing on that interval. If $\phi'(x) < 0$ on an interval, then $\phi(x)$ is decreasing on that interval.
• Second Derivative and Monotonicity of the First Derivative: If $f''(x) > 0$ on an interval, then $f'(x)$ is an increasing function on that interval.
• Chain Rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

Step-by-Step Solution

1. Understand the Given Information and Its Implications

We are given that $f: [0, 2] \to \mathbb{R}$ is twice differentiable and $f''(x) > 0$ for all $x \in (0, 2)$. This implies that $f'(x)$ is an increasing function on the interval $(0, 2)$. That is, if $a < b$ for $a, b \in (0, 2)$, then $f'(a) < f'(b)$. This will be crucial for determining the monotonicity of $\phi(x)$. We are also given $\phi(x) = f(x) + f(2 - x)$. Our goal is to determine whether $\phi(x)$ is increasing or decreasing on the interval $(0, 2)$.

2. Calculate the First Derivative of $\phi(x)$

To analyze the monotonicity of $\phi(x)$, we need to find $\phi'(x)$. $\phi(x) = f(x) + f(2 - x)$ Differentiating with respect to $x$: $\phi'(x) = \frac{d}{dx} [f(x) + f(2 - x)]$ Using the sum rule for differentiation: $\phi'(x) = \frac{d}{dx} f(x) + \frac{d}{dx} f(2 - x)$ The derivative of $f(x)$ is $f'(x)$. Using the chain rule for $f(2 - x)$: Let $u = 2 - x$, so $\frac{du}{dx} = -1$. Then, $\frac{d}{dx} f(2 - x) = f'(u) \cdot \frac{du}{dx} = f'(2 - x) \cdot (-1) = -f'(2 - x)$. Therefore: $\phi'(x) = f'(x) - f'(2 - x)$ This expression for $\phi'(x)$ will allow us to determine the intervals where $\phi(x)$ is increasing or decreasing using the First Derivative Test.

3. Analyze the Sign of $\phi'(x)$ to Determine Monotonicity

We want to find where $\phi'(x) > 0$ (increasing) and $\phi'(x) < 0$ (decreasing). We know $f'(x)$ is an increasing function because $f''(x) > 0$.

Case 1: $\phi(x)$ is increasing For $\phi(x)$ to be increasing, we need $\phi'(x) > 0$: $f'(x) - f'(2 - x) > 0$ $f'(x) > f'(2 - x)$ Since $f'(x)$ is an increasing function, $f'(a) > f'(b)$ implies $a > b$. Therefore: $x > 2 - x$ $2x > 2$ $x > 1$ So, $\phi(x)$ is increasing on the interval $(1, 2)$.

Case 2: $\phi(x)$ is decreasing For $\phi(x)$ to be decreasing, we need $\phi'(x) < 0$: $f'(x) - f'(2 - x) < 0$ $f'(x) < f'(2 - x)$ Since $f'(x)$ is an increasing function, $f'(a) < f'(b)$ implies $a < b$. Therefore: $x < 2 - x$ $2x < 2$ $x < 1$ So, $\phi(x)$ is decreasing on the interval $(0, 1)$.

4. Combine the Results

We have found that:

• $\phi(x)$ is decreasing on $(0, 1)$.
• $\phi(x)$ is increasing on $(1, 2)$.

This matches option (B).

Summary and Key Takeaway

The problem involves analyzing the monotonicity of a composite function $\phi(x)$ using the properties of its derivatives. The key is to recognize that $f''(x) > 0$ implies that $f'(x)$ is increasing, and to use this information to determine the sign of $\phi'(x)$. The function $\phi(x)$ is decreasing on $(0, 1)$ and increasing on $(1, 2)$, which corresponds to option (B).

Common Mistakes & Tips

1. Sign Errors with the Chain Rule: When differentiating $f(2-x)$, remember the chain rule introduces a negative sign.
2. Misinterpreting $f''(x) > 0$: This means $f'(x)$ is increasing. If you incorrectly assume $f'(x)$ is decreasing, you will reverse the inequalities and get the wrong answer.
3. Not Using the Increasing Property of $f'(x)$ Correctly: The fact that $f'(x)$ is increasing is crucial. $f'(a) > f'(b)$ implies $a > b$ because $f'$ is increasing.

Final Answer: The final answer is $\boxed{\text{(B)}}$, decreasing on (0, 1) and increasing on (1, 2).