To determine whether the given function can be a probability density function (PDF) of a continuous random variable, we need to verify two conditions:

1. Non-negativity: $f(x) \geq 0$ for all $x$ in the domain.
2. Normalization: The total area under the curve of $f(x)$ over the entire real line must equal 1. Mathematically, this is expressed as: $\int_{-\infty}^{\infty} f(x) \, dx = 1$

Given the function $f(x)$ in the image:

x, & -1 \leq x \leq 1 \\ 0, & \text{otherwise} \end{cases}$$ ### Step 1: Check Non-Negativity The function $$f(x)$$ is defined as $$x$$ within the interval $$[-1, 1]$$. In this interval: - $$f(x)$$ is non-negative because $$-1 \leq x \leq 1$$, and the value of $$x$$ is non-negative for $$0 \leq x \leq 1$$. Outside this interval, $$f(x) = 0$$, which is also non-negative. Thus, the non-negativity condition is satisfied. ### Step 2: Check Normalization We need to calculate the integral of $$f(x)$$ over the interval $$[-1, 1]$$: $$\int_{-\infty}^{\infty} f(x) \, dx = \int_{-1}^{1} x \, dx$$ The integral of $$x$$ with respect to $$x$$ is: $$\int_{-1}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{1} = \frac{(1)^2}{2} - \frac{(-1)^2}{2} = \frac{1}{2} - \frac{1}{2} = 0$$ Since the integral equals 0, this function does **not** satisfy the normalization condition required for a PDF (which must integrate to 1 over its domain). ### Conclusion The given function **cannot** be the probability density function of a continuous random variable because it does not satisfy the normalization condition. Do you want more details, or have any questions? Here are some related questions: 1. How do you modify a function to make it a valid probability density function? 2. What are some examples of valid probability density functions? 3. How is the concept of a cumulative distribution function (CDF) related to the PDF? 4. What does it mean when the integral of a function is zero? 5. Can a function with a negative integral be a probability density function? **Tip:** For any function to qualify as a probability density function, always check that its integral over the entire space equals 1, indicating that it covers the whole probability space.