find the domain and range of the following functions h(x) { 2x +1, if x >0     x²       if x≥0

To find the domain and range of the piecewise function \( h(x) \) defined as follows:

\[ h(x) = \begin{cases} 
2x + 1 & \text{if } x > 0 \\
x^2 & \text{if } x \ge 0 
\end{cases} \]

### Domain
The domain of a function is the set of all possible input values (x-values) for which the function is defined. 

In this case:

- For \( 2x + 1 \), the function is defined for \( x > 0 \).
- For \( x^2 \), the function is defined for \( x \ge 0 \).

Since both parts cover all non-negative real numbers, the domain of \( h(x) \) is:

\[ \text{Domain} = [0, \infty) \]

### Range
The range of a function is the set of all possible output values (y-values) that the function can produce.

Consider each piece separately:

1. For \( x^2 \) when \( x \ge 0 \):
   - The smallest value of \( x^2 \) is 0 (when \( x = 0 \)).
   - As \( x \) increases, \( x^2 \) increases without bound.

   Therefore, the range of \( x^2 \) for \( x \ge 0 \) is \([0, \infty)\).

2. For \( 2x + 1 \) when \( x > 0 \):
   - When \( x \) is just greater than 0, \( 2x + 1 \) is just greater than 1.
   - As \( x \) increases, \( 2x + 1 \) also increases without bound.

   Therefore, the range of \( 2x + 1 \) for \( x > 0 \) is \((1, \infty)\).

To combine these ranges, we note that the output from the first part is from 0 upwards, and from the second part is from just above 1 upwards. Hence, the combined range is:

\[ \text{Range} = [0, \infty) \]

### Summary
- **Domain**: \([0, \infty)\)
- **Range**: \([0, \infty)\)

Would you like more details on this topic or have any questions?

Here are some questions you might be interested in:
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5

Learn how to find the domain and range of the piecewise function h(x) = { 2x + 1, if x > 0; x^2, if x ≥ 0 }. This problem explores the concept of piecewise functions, domain, and range in mathematics. Suitable for Grades 10-12.

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