Final Answer:
The following density function describes a random variable x. The value of f(x) for 0 < x < 2 is 3x^2.
Step-by-step explanation:
The given density function is:
f(x) =
begin{cases}
3x^2, & 0 < x < 2 \\
0, & \text{otherwise}
\end{cases}
To find the value of f(x) for 0 < x < 2, we simply substitute the value of x into the expression for f(x) in this range. Therefore, we have:
f(x) = 3x^2, for 0 < x < 2.
This function is a quadratic function of x, where the coefficient of x^2 is 3. The graph of this function is a parabola that opens upwards, with its vertex at the origin (0,0). The maximum value of the function occurs at x=1, where f(1) = 3. As x approaches 0 or 2 from within the range [0,2], the value of f(x) approaches zero. This behavior is typical of a density function, which integrates to one over the range of interest. In this case, the integral of f(x) over [0,2] is:
∫_0^2 3x^2 dx = [x^3]_0^2 = 8. This confirms that f(x) is indeed a valid density function.