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The problem you've uploaded provides the following information:

Given a positive real-valued function \( f \), which is differentiable for all \( x \in \mathbb{R} \), and also satisfies:
- \( f(0) = 0 \)
- \( f(1) = 3 \),

The function \( g(x) \) is defined as:
\[
g(x) = \int_0^x f(t) \, dt
\]

Then, there are multiple options regarding \( g(x) \) to evaluate:

1. \( g \) has an inflection point at \( x = 1 \).
2. \( g \) has a local maximum at \( x = 1 \).
3. \( g(1) = f(1) - f\left( \frac{1}{2} \right) \).
4. \( g \) has a local minimum at \( x = 1 \).

### Analysis:

1. **Inflection point at \( x = 1 \)**: For an inflection point, the second derivative of \( g(x) \) must change sign at \( x = 1 \). The second derivative of \( g(x) \) is \( f'(x) \), so this would require checking the behavior of \( f'(x) \) around \( x = 1 \).

2. **Local maximum at \( x = 1 \)**: For a local maximum, we need \( g'(x) = 0 \) and \( g''(x) < 0 \). The derivative \( g'(x) = f(x) \), so for \( g \) to have a local maximum, \( f(1) = 0 \), but from the problem, \( f(1) = 3 \), so this option is false.

3. **\( g(1) = f(1) - f\left( \frac{1}{2} \right) \)**: This option provides a specific expression for \( g(1) \), which can be verified by calculating the integral of \( f(t) \) from 0 to 1.

4. **Local minimum at \( x = 1 \)**: Similar to the local maximum case, we would need to check \( f'(x) \). Since \( f(1) = 3 \), this implies \( g'(1) = 3 \), so \( g(x) \) does not have a local minimum at \( x = 1 \).

I can analyze this further by computing the values and derivatives of \( g(x) \) for more precise verification. Would you like me to proceed with this?

Given a positive real-valued differentiable function f(x), and g(x) = ∫ f(t) dt from 0 to x, analyze the characteristics of g at x=1.

Explore the analysis of the function g(x), derived from the integral of a differentiable function f(x), and the behavior of g at x=1. This problem involves key calculus concepts such as derivatives, definite integrals, and critical points.

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