Question

Let g(x) = 2f(x/2)+f(1-x) and f"(x) < 0 in 0 ≤ x ≤ 1 then g(x)

• A. decreases in $\left[0, \frac{2}{3}\right]$
• B. decreases $\left[\frac{2}{3}, 1\right]$
• C. increases in $\left[0, \frac{2}{3}\right]$
• D. increases in $\left[\frac{2}{3}, 1\right]$

Answer 1

Answer: (B), (C)

decreases $\left[\frac{2}{3}, 1\right]$, increases in $\left[0, \frac{2}{3}\right]$

Answer 2

The function given is $g(x) = 2f(x/2) + f(1-x)$ for $x \in [0, 1]$.
We are given that $f''(x) < 0$ for $x \in [0, 1]$. This implies that the first derivative $f'(x)$ is a decreasing function in the interval $[0, 1]$.

To determine the monotonicity of $g(x)$, we need to find the sign of its derivative $g'(x)$.
Differentiating $g(x)$ with respect to $x$, we use the chain rule:
$g'(x) = \frac{d}{dx}(2f(x/2)) + \frac{d}{dx}(f(1-x))$
$g'(x) = 2 \cdot f'(x/2) \cdot \frac{d}{dx}(x/2) + f'(1-x) \cdot \frac{d}{dx}(1-x)$
$g'(x) = 2 \cdot f'(x/2) \cdot \frac{1}{2} + f'(1-x) \cdot (-1)$
$g'(x) = f'(x/2) - f'(1-x)$

Now, we need to determine the sign of $g'(x)$ for $x \in [0, 1]$. The sign depends on the comparison between $f'(x/2)$ and $f'(1-x)$.
Since $f'(x)$ is a decreasing function, $f'(a) > f'(b)$ if $a < b$, and $f'(a) < f'(b)$ if $a > b$.

Let's compare the arguments of $f'$: $x/2$ and $1-x$.
We find the point where the arguments are equal:
$x/2 = 1-x$
$x = 2(1-x)$
$x = 2 - 2x$
$3x = 2$
$x = 2/3$.

At $x = 2/3$, $x/2 = 1/3$ and $1-x = 1/3$. So $f'(x/2) = f'(1-x)$, which means $g'(2/3) = 0$.

Now consider the intervals $x < 2/3$ and $x > 2/3$ within the domain $[0, 1]$.

Case 1: $x \in [0, 2/3)$
For $x < 2/3$, we compare $x/2$ and $1-x$.
$x/2 < 1-x \iff x < 2 - 2x \iff 3x < 2 \iff x < 2/3$.
So, for $x \in [0, 2/3)$, we have $x/2 < 1-x$.
Since $f'(x)$ is a decreasing function, if $a < b$, then $f'(a) > f'(b)$.
Here, $a = x/2$ and $b = 1-x$, with $a < b$.
So, $f'(x/2) > f'(1-x)$.
Therefore, $g'(x) = f'(x/2) - f'(1-x) > 0$ for $x \in [0, 2/3)$.

Case 2: $x \in (2/3, 1]$
For $x > 2/3$, we compare $x/2$ and $1-x$.
$x/2 > 1-x \iff x > 2 - 2x \iff 3x > 2 \iff x > 2/3$.
So, for $x \in (2/3, 1]$, we have $x/2 > 1-x$.
Since $f'(x)$ is a decreasing function, if $a > b$, then $f'(a) < f'(b)$.
Here, $a = x/2$ and $b = 1-x$, with $a > b$.
So, $f'(x/2) < f'(1-x)$.
Therefore, $g'(x) = f'(x/2) - f'(1-x) < 0$ for $x \in (2/3, 1]$.

Combining the results:
$g'(x) > 0$ for $x \in [0, 2/3)$ and $g'(x) < 0$ for $x \in (2/3, 1]$.
At $x = 2/3$, $g'(2/3) = 0$.
This means $g(x)$ is increasing on the interval $[0, 2/3]$ and decreasing on the interval $[2/3, 1]$.

The options are multiple choice checkboxes, indicating multiple correct answers are possible.
Based on our analysis, options 2 and 3 are correct.