Function Discontinuities 15Fe43

1. **State the problem:** We are given a function $y = f(x)$ with discontinuities at $x = -1$, $x = 0$, and $x = 1$. We need to find these discontinuities, explain why the formal definition of continuity does not apply at these points, and classify each discontinuity. 2. **Recall the definition of continuity:** A function $f$ is continuous at $x = a$ if: - $f(a)$ is defined. - The limit $\lim_{x \to a} f(x)$ exists. - $\lim_{x \to a} f(x) = f(a)$. 3. **Analyze each discontinuity:** - At $x = -1$: - The graph shows an open circle at $(-1, 1)$ and a filled point at $(-1, 3)$. - So, $f(-1) = 3$ but the limit from the left and right approaches $1$. - The limit $\lim_{x \to -1} f(x) = 1$ but $f(-1) = 3$, so the limit does not equal the function value. - This is a **jump discontinuity** because the left and right limits exist but are not equal to the function value. - At $x = 0$: - There is an open circle at $(0, 3)$ but no filled point shown. - This means $f(0)$ is not defined. - Since $f(0)$ is undefined, the function is discontinuous at $x=0$. - This is a **removable discontinuity** because the limit exists but the function is not defined at that point. - At $x = 1$: - There is an open circle at $(1, 1)$ and a filled point at $(1, 2)$. - So, $f(1) = 2$ but the limit from the left and right approaches $1$. - The limit $\lim_{x \to 1} f(x) = 1$ but $f(1) = 2$, so the limit does not equal the function value. - This is a **jump discontinuity**. 4. **Summary:** | $x$ | Discontinuity Type | Reason | |------|--------------------|--------| | $-1$ | Jump | Limit exists but $\lim f(x) \neq f(-1)$ | | $0$ | Removable | $f(0)$ undefined but limit exists | | $1$ | Jump | Limit exists but $\lim f(x) \neq f(1)$ | **Final answers:** - Discontinuities at $x = -1, 0, 1$. - At $x = -1$ and $x = 1$, the function has jump discontinuities. - At $x = 0$, the function has a removable discontinuity because $f(0)$ is not defined.