Main Question

A piecewise function is a function defined by different expressions for different intervals of the domain.

Evaluating the function:

Given:

$$f(x) = \begin{cases} 1 + x & \text{if } x \leq -1 \\ x^2 & \text{if } x > -1 \end{cases}$$

a) Finding $f(-3)$:

Since $-3 \leq -1$, we use the first piece: $f(x) = 1 + x$

$$f(-3) = 1 + (-3) = \boxed{-2}$$

b) Finding $f(-1)$:

Since $-1 \leq -1$, we use the first piece: $f(x) = 1 + x$

$$f(-1) = 1 + (-1) = \boxed{0}$$

c) Finding $f(0)$:

Since $0 > -1$, we use the second piece: $f(x) = x^2$

$$f(0) = (0)^2 = \boxed{0}$$

Sketch of the Graph:

Second Question

Discontinuity at a point occurs when a function is not continuous at that point, i.e., $\lim_{x \to a} f(x) \neq f(a)$ or the limit does not exist.

Types of Discontinuity:

• Removable Discontinuity: The limit $\lim_{x \to a} f(x)$ exists but is either not equal to $f(a)$ or $f(a)$ is not defined. The "hole" can be removed by redefining $f(a)$.

• Jump Discontinuity: The left-hand limit and right-hand limit both exist but are not equal, i.e., $\lim_{x \to a^-} f(x) \neq \lim_{x \to a^+} f(x)$.

• Infinite Discontinuity: At least one of the one-sided limits tends to $\pm\infty$. This occurs at vertical asymptotes.

Analysing the given function:

$$f(x) = \frac{x-2}{x^2 - 7x + 10}$$

Step 1: Factorize the denominator:

$$x^2 - 7x + 10 = (x-2)(x-5)$$

Step 2: Simplify the function:

$$f(x) = \frac{x-2}{(x-2)(x-5)} = \frac{1}{x-5}, \quad x \neq 2$$

Step 3: Identify points of discontinuity:

The function is undefined at $x = 2$ and $x = 5$.

At $x = 2$ (Removable Discontinuity):

$$\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{1}{x-5} = \frac{1}{2-5} = \frac{-1}{3}$$

The limit exists but $f(2)$ is undefined. So this is a removable discontinuity.

At $x = 5$ (Infinite Discontinuity):

$$\lim_{x \to 5} \frac{1}{x-5} = \pm\infty$$

The function tends to infinity, so this is an infinite discontinuity (vertical asymptote).

Conclusion:

The function is discontinuous at $x = 2$ and $x = 5$. It is continuous for all $x \in \mathbb{R} \setminus \{2, 5\}$, i.e., at every point except $x = 2$ (removable) and $x = 5$ (infinite).