$$ y = \left\{ \begin{array} { l l } { x ^ { 2 } , } & { x 0 } \ { x - 2 , } & { x \leq 0 } \end{array} ight. $$

Question

$$ y = \left\{ \begin{array} { l l } { x ^ { 2 } , } & { x > 0 } \ { x - 2 , } & { x \leq 0 } \end{array} ight. $$

UpStudy AI · Accepted Answer

You've defined a **piecewise function** $$ y $$ that behaves differently depending on the value of $$ x $$. Let's break it down and analyze its properties.

### The Function

$$
y = \begin{cases} 
x^{2} & \text{if } x > 0, \
x - 2 & \text{if } x \leq 0.
\end{cases}
$$

### Graphical Representation

To understand the function better, let's consider its graph:

1. **For $$ x > 0 $$:**
   - The function is $$ y = x^{2} $$, which is a standard upward-opening parabola.
   
2. **For $$ x \leq 0 $$:**
   - The function is $$ y = x - 2 $$, which is a straight line with a slope of 1 and a y-intercept at $$-2$$.

Here's a rough sketch of the graph:

```
          |
        4 |         *
          |       *
        2 |     *
          |   *
        0 |---*--------- 
          |  /
       -2 | /
          |/
          +----------------
```

- The parabola $$ y = x^{2} $$ starts at $$ x = 0 $$ and extends to the right.
- The line $$ y = x - 2 $$ starts at $$ x = 0 $$ and extends to the left.

### Continuity at $$ x = 0 $$

A function is **continuous** at a point if the following three conditions are met:
1. The function is defined at that point.
2. The limit exists at that point.
3. The limit equals the function's value at that point.

Let's check continuity at $$ x = 0 $$:

1. **Function value at $$ x = 0 $$:**
   $$
   y(0) = 0 - 2 = -2
   $$

2. **Left-hand limit as $$ x $$ approaches 0 (from the left):**
   $$
   \lim_{x \to 0^-} y = \lim_{x \to 0^-} (x - 2) = -2
   $$

3. **Right-hand limit as $$ x $$ approaches 0 (from the right):**
   $$
   \lim_{x \to 0^+} y = \lim_{x \to 0^+} x^{2} = 0
   $$

Since the left-hand limit ($$-2$$) does not equal the right-hand limit ($$0$$), the function $$ y $$ is **discontinuous** at $$ x = 0 $$.

### Differentiability

A function must be continuous at a point to be differentiable there. Since $$ y $$ is discontinuous at $$ x = 0 $$, it is **not differentiable** at that point.

### Summary of Properties

- **Domain:** All real numbers ($$ \mathbb{R} $$)
- **Range:** $$ y \geq -2 $$
- **Continuity:** Discontinuous at $$ x = 0 $$
- **Differentiability:** Not differentiable at $$ x = 0 $$, but differentiable elsewhere
- **Behavior:**
  - For $$ x > 0 $$, $$ y $$ grows quadratically.
  - For $$ x \leq 0 $$, $$ y $$ decreases linearly.

### Applications

Piecewise functions like this are useful in modeling real-world scenarios where a system behaves differently under different conditions. For example:

- **Tax Brackets:** Tax rates change at different income levels.
- **Engineering:** Material properties can change under different temperature ranges.
- **Economics:** Different pricing strategies may apply beyond certain sales volumes.

If you have a specific question or need further analysis on this function, feel free to ask!
The function $$ y $$ is defined as: $$ y = \begin{cases} x^{2} & \text{if } x > 0, \ x - 2 & \text{if } x \leq 0. \end{cases} $$ It is a piecewise function with a parabola for $$ x > 0 $$ and a straight line for $$ x \leq 0 $$. The function is discontinuous at $$ x = 0 $$ and not differentiable there.

\( y = \left\{ \begin{array} { l l } { x ^ { 2 } , } & { x > 0 } \\ { x - 2 , } & { x \leq 0 } \end{array} \right. \)

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