1. **State the problem:** We need to find the value of $x$ where the piecewise function $$f(x) = \begin{cases} 5x, & x < 0 \\ 1, & x = 0 \\ -5x, & x > 0 \end{cases}$$ is discontinuous. 2. **Recall the definition of continuity at a point $x = a$:** A function $f$ is continuous at $x = a$ if $$\lim_{x \to a^-} f(x) = f(a) = \lim_{x \to a^+} f(x).$$ 3. **Check continuity at $x=0$ (the only point where the definition changes):** - Left-hand limit: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 5x = 5 \times 0 = 0.$$ - Right-hand limit: $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} -5x = -5 \times 0 = 0.$$ - Function value: $$f(0) = 1.$$ 4. **Compare limits and function value:** $$\lim_{x \to 0^-} f(x) = 0 \neq 1 = f(0),$$ so the function is discontinuous at $x=0$. 5. **Check other points:** For $x<0$ and $x>0$, the function is linear and continuous. **Final answer:** The function is discontinuous at $x=0$.