AMPLE 3 en: $$ f(x)=2 x^{2} $$ Determine the equation of the inverse of $$ f $$ in the form $$ f^{-1}(x)=\ldots $$ On the same set of $$ x $$-axis, sketch the graphs of $$ f $$ and $$ f^{-1} $$. Indicate clearly the intercepts with the axis, as well as another point on the graph of each f and $$ f^{-1} $$. Determine the domain and range of both $$ f $$ and $$ f^{-1} $$. Is $$ f^{-1} $$ a function? Give a reason for your answer. How can $$ f $$ be restricted so that the inverse is a function? On the separate set of axis, sketch both restricted function and their inverses.

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It seems that I am unable to generate plots directly. However, I can guide you on how to sketch the graphs of $$ f(x) = 2x^2 $$ and its inverse $$ f^{-1}(x) = \sqrt{\frac{x}{2}} $$.

### Instructions for Sketching the Graphs

1. **Graph of $$ f(x) = 2x^2 $$**:
   - Start at the origin (0, 0).
   - For $$ x = 1 $$, $$ f(1) = 2 $$ (point (1, 2)).
   - For $$ x = 2 $$, $$ f(2) = 8 $$ (point (2, 8)).
   - The graph is a parabola opening upwards, starting from the origin and increasing steeply as $$ x $$ increases.

2. **Graph of $$ f^{-1}(x) = \sqrt{\frac{x}{2}} $$**:
   - Start at the origin (0, 0).
   - For $$ x = 2 $$, $$ f^{-1}(2) = 1 $$ (point (2, 1)).
   - For $$ x = 8 $$, $$ f^{-1}(8) = 2 $$ (point (8, 2)).
   - The graph is a curve that increases, starting from the origin and gradually increasing, but at a slower rate than the parabola.

### Key Points to Indicate on the Graphs
- **For $$ f(x) $$**:
  - Intercept: (0, 0)
  - Point: (1, 2)

- **For $$ f^{-1}(x) $$**:
  - Intercept: (0, 0)
  - Point: (2, 1)

### Domain and Range
- **Both functions** have the same domain and range:
  - Domain: $$ [0, \infty) $$
  - Range: $$ [0, \infty) $$

### Conclusion
- The graphs of $$ f $$ and $$ f^{-1} $$ will reflect across the line $$ y = x $$, which is a characteristic of inverse functions.

If you have graphing software or graph paper, you can plot these points and draw the curves accordingly. If you need further assistance or have any questions, feel free to ask!
To find the inverse of $$ f(x) = 2x^2 $$, restrict the domain to $$ x \geq 0 $$, resulting in $$ f^{-1}(x) = \sqrt{\frac{x}{2}} $$. Both functions have the same domain and range of $$ [0, \infty) $$. The inverse is a function because it passes the vertical line test. To make the inverse a function, restrict $$ f $$ to $$ x \geq 0 $$.

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