Questions: c) h(x)=3x/(x^2+3) → Calcular su dominio recorrido

Solution Steps

To find the domain of the function \( h(x) = \frac{3x}{x^2 + 3} \), we need to determine the values of \( x \) for which the function is defined. Since the denominator \( x^2 + 3 \) is always positive for all real numbers, the function is defined for all real \( x \). Therefore, the domain is all real numbers. To find the range, we need to analyze the behavior of the function as \( x \) approaches positive and negative infinity and any critical points.

The function \( h(x) = \frac{3x}{x^2 + 3} \) is defined for all real numbers because the denominator \( x^2 + 3 \) is always positive for any real \( x \). Therefore, the domain of \( h(x) \) is all real numbers.

To find the critical points, we take the derivative of \( h(x) \) and set it to zero:

\[ h'(x) = \frac{d}{dx} \left( \frac{3x}{x^2 + 3} \right) \]

Solving \( h'(x) = 0 \) gives the critical points \( x = -\sqrt{3} \) and \( x = \sqrt{3} \).

To determine the behavior of the function as \( x \) approaches infinity, we calculate the limits:

\[ \lim_{x \to \infty} h(x) = 0 \quad \text{and} \quad \lim_{x \to -\infty} h(x) = 0 \]

This indicates that the function approaches 0 as \( x \) goes to positive or negative infinity.

Evaluate the function at the critical points:

\[ h(-\sqrt{3}) = -\frac{\sqrt{3}}{2} \quad \text{and} \quad h(\sqrt{3}) = \frac{\sqrt{3}}{2} \]

Considering the behavior at infinity and the values at the critical points, the range of the function is:

\[ \left(-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}\right) \]

Final Answer