Let's break down the problem step by step based on the image you provided.

We are given the following information:

• $f(x)$ is positive and concave up on the interval $I$.
• $f''(x)$ exists, and $g(x) = (f(x))^2$.

We are asked to choose from a list to answer specific questions about $f(x)$, $g(x)$, and their derivatives on the interval $I$. The choices are: CU (concave up), CD (concave down), $f(x)$, $f'(x)$, $f''(x)$, 0, or 1.

a) $f''(x) > 0$ on $I$

Since we know that $f(x)$ is concave up on $I$, this directly implies that $f''(x) > 0$ on $I$.

Answer for a): CU (concave up).

b) $g'(x) = 2(A^2 + B f''(x))$, where $A = f'(x)$ and $B = f(x)$

First, we need to find $g'(x)$, which is the derivative of $g(x) = (f(x))^2$.

Using the chain rule: $g'(x) = 2 f(x) f'(x)$ So, this simplifies to $g'(x) = 2 f(x) f'(x)$. Now, we look at the information we have:

• $f(x)$ is positive.
• $f'(x)$ is likely positive because $f(x)$ is increasing if concave up.

Thus, $g'(x) > 0$.

Answer for b): 0.

c) $g''(x) > 0$ on $I$

Now, we find $g''(x)$, which is the derivative of $g'(x) = 2 f(x) f'(x)$.

Using the product rule: $g''(x) = 2(f'(x) f'(x) + f(x) f''(x)) = 2((f'(x))^2 + f(x) f''(x))$ Since $f(x) > 0$, $f'(x) \geq 0$, and $f''(x) > 0$, we conclude that $g''(x) > 0$.

Answer for c): CU (concave up).

d) $g(x)$ is __ on $I$

Given that $g(x) = (f(x))^2$ and $f(x) > 0$, $g(x)$ is increasing and concave up.

Answer for d): CU (concave up).

Do you need further clarification or additional details?

Here are 5 follow-up questions for practice:

1. How does concavity relate to the second derivative?
2. Can a function be concave up while its derivative is decreasing?
3. What are the conditions for $g'(x) = 0$?
4. How would the behavior change if $f(x)$ were concave down?
5. How can you determine inflection points from the second derivative?

Tip: When analyzing a function’s concavity, always look at the sign of the second derivative!