Consider the curves
`f(x)=x|x|-1andg(x)={{:((3x)/(2)","xgt0),(2x","xle0):}`
Where do the curves intersect?

Options

A. At(2,3) only
B. At (-1,-2) only
C. At (2,3)and(-1,-2)
D. Neither at (2,3)nor at (-1,-2)

Explanation

To find the points of intersection of the curves \( f(x) = x|x| - 1 \) and \( g(x) = \begin{cases} \frac{3x}{2} & \text{if } x > 0 \\ 2x & \text{if } x \leq 0 \end{cases} \), we will follow these steps:

### Step 1: Define the functions
We start with the given functions:
- For \( f(x) \):
\[ f(x) = x|x| - 1 \]
This can be rewritten based on the value of \( x \):
- If \( x \geq 0 \): \( f(x) = x^2 - 1 \)
- If \( x < 0 \): \( f(x) = -x^2 - 1 \)

- For \( g(x) \):
\[ g(x) = \begin{cases} \frac{3x}{2} & \text{if } x > 0 \\ 2x & \text{if } x \leq 0 \end{cases} \]

### Step 2: Find intersections for \( x \geq 0 \)
For \( x \geq 0 \):
\[ f(x) = x^2 - 1 \quad \text{and} \quad g(x) = \frac{3x}{2} \]
Setting these equal to find the intersection:
\[ x^2 - 1 = \frac{3x}{2} \]
Multiplying through by 2 to eliminate the fraction:
\[ 2x^2 - 2 = 3x \]
Rearranging gives:
\[ 2x^2 - 3x - 2 = 0 \]
Now we will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -3, c = -2 \):
\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} \]
Calculating the discriminant:
\[ x = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm \sqrt{25}}{4} = \frac{3 \pm 5}{4} \]
This gives us two solutions:
\[ x = \frac{8}{4} = 2 \quad \text{and} \quad x = \frac{-2}{4} = -\frac{1}{2} \quad (\text{not valid for } x \geq 0) \]
Thus, for \( x \geq 0 \), the intersection point is \( x = 2 \).

Calculating \( y \) at this point:
\[ f(2) = 2^2 - 1 = 3 \quad \text{and} \quad g(2) = \frac{3 \cdot 2}{2} = 3 \]
So the intersection point is \( (2, 3) \).

### Step 3: Find intersections for \( x < 0 \)
For \( x < 0 \):
\[ f(x) = -x^2 - 1 \quad \text{and} \quad g(x) = 2x \]
Setting these equal:
\[ -x^2 - 1 = 2x \]
Rearranging gives:
\[ -x^2 - 2x - 1 = 0 \quad \text{or} \quad x^2 + 2x + 1 = 0 \]
Factoring:
\[ (x + 1)^2 = 0 \]
This gives:
\[ x = -1 \]
Calculating \( y \) at this point:
\[ f(-1) = -(-1)^2 - 1 = -1 - 1 = -2 \quad \text{and} \quad g(-1) = 2 \cdot (-1) = -2 \]
So the intersection point is \( (-1, -2) \).

### Conclusion
The curves intersect at the points \( (2, 3) \) and \( (-1, -2) \).

### Final Answer
The points of intersection are:
1. \( (2, 3) \)
2. \( (-1, -2) \)