To determine the nature of the function \( f: \mathbb{R} \to \mathbb{R} \) defined by \(f(x) = \begin{cases} 2x+3, & x \le \frac{4}{3} \\ -3x^2+8x, & x > \frac{4}{3} \end{cases}\), let's analyze each segment.
Step 1: Analyze the segment \( f(x) = 2x + 3 \) for \( x \le \frac{4}{3} \)
- \(f(x) = 2x + 3\) is a linear function. A linear function with a non-zero coefficient of \( x \) is one-to-one unless restricted to a domain making it constant. Here, it's defined on the domain \( x \le \frac{4}{3} \), so it remains one-to-one.
Step 2: Analyze the segment \( f(x) = -3x^2 + 8x \) for \( x > \frac{4}{3} \)
- \(f(x) = -3x^2 + 8x\) is a quadratic function opening downwards because the coefficient of \( x^2 \) is negative. Quadratic functions are not one-to-one unless specifically restricted in their domain.
Step 3: Determine if the complete function is onto (surjective)
- A function \( f: \mathbb{R} \to \mathbb{R} \) is onto if for every real number \( y \), there exists some real \( x \) such that \( f(x) = y \).
- For \( x \le \frac{4}{3} \), the range is \([-\infty, \frac{11}{3}]\).
- For \( x > \frac{4}{3} \), since \( -3x^2 + 8x \) is a downward opening parabola with vertex at \(x = \frac{4}{3}\), it will take maximum values at the turning point and tend towards negative infinity.
- An analysis of the vertex and the direction of the parabola indicates that for \( x > \frac{4}{3} \), there are gaps in the real numbers that cannot be reached by \( f(x) \).
Conclusion: The function is not onto because it does not cover all real numbers in its range. Thus, the correct answer is "not onto".