Find the inverse of the function: `f(x)={x^3-1, ,x<2x^2+3,xgeq2`

Explanation

To find the inverse of the given piecewise function \( f(x) \), we need to consider each piece of the function separately.

### Step 1: Identify the pieces of the function
The function is defined as:
\[ f(x) = \begin{cases} x^3 - 1 & \text{if } x < 2 \\ 2x^2 + 3 & \text{if } x \geq 2 \end{cases} \]

### Step 2: Find the inverse for the first piece \( f(x) = x^3 - 1 \) when \( x < 2 \)
1. Set \( y = f(x) \):
\[ y = x^3 - 1 \]
2. Solve for \( x \):
\[ y + 1 = x^3 \implies x = (y + 1)^{1/3} \]
3. Since this piece is defined for \( x < 2 \), we need to determine the range of \( f(x) \) for \( x < 2 \):
- When \( x \to 2 \), \( f(x) \to 2^3 - 1 = 7 \).
- As \( x \) approaches negative infinity, \( f(x) \) approaches negative infinity.
- Therefore, the range for this piece is \( (-\infty, 7) \).

Thus, the inverse for this piece is:
\[ f^{-1}(x) = (x + 1)^{1/3} \quad \text{for } x < 7 \]

### Step 3: Find the inverse for the second piece \( f(x) = 2x^2 + 3 \) when \( x \geq 2 \)
1. Set \( y = f(x) \):
\[ y = 2x^2 + 3 \]
2. Solve for \( x \):
\[ y - 3 = 2x^2 \implies x^2 = \frac{y - 3}{2} \implies x = \sqrt{\frac{y - 3}{2}} \]
(We take the positive root since \( x \geq 2 \)).
3. Determine the range of \( f(x) \) for \( x \geq 2 \):
- When \( x = 2 \), \( f(x) = 2(2^2) + 3 = 11 \).
- As \( x \) approaches positive infinity, \( f(x) \) approaches positive infinity.
- Therefore, the range for this piece is \( [11, \infty) \).

Thus, the inverse for this piece is:
\[ f^{-1}(x) = \sqrt{\frac{x - 3}{2}} \quad \text{for } x \geq 11 \]

### Final Result
Combining both pieces, we have the inverse function:
\[ f^{-1}(x) = \begin{cases} (x + 1)^{1/3} & \text{if } x < 7 \\ \sqrt{\frac{x - 3}{2}} & \text{if } x \geq 11 \end{cases} \]