The range of ℎ(𝑥)={2𝑥+1𝑥<13𝑥≥1h(x)={ 2x+13​ x<1x≥1​ is:A.(−∞,3](−∞,3]B.[ 3,∞∞)C.(−∞,3)(−∞,3)D.(−∞,1)(−∞,1)E.All real numbersF.(−∞,1](−∞,1]

1. Break Down the Problem

We need to find the range of the piecewise function $h(x)$, which is defined as follows: $h(x) = \begin{cases} 2x + 1 & \text{if } x < 1 \\ 3 & \text{if } x \geq 1 \end{cases}$ We will analyze the two cases separately to determine their ranges.

2. Relevant Concepts

The range of a function is the set of possible output values. To find the range, we need to evaluate the function for both cases of $x$:

1. For $x < 1$: Evaluate $h(x) = 2x + 1$.
2. For $x \geq 1$: Evaluate $h(x) = 3$.

3. Analysis and Detail

Case 1: $x < 1$

• The function is $h(x) = 2x + 1$.
• As $x$ approaches $1$ from the left, $h(x)$ approaches $2(1) + 1 = 3$.
• As $x$ decreases towards negative infinity, $h(x)$ decreases without bound. Thus, we analyze $h(x)$:

$\lim_{x \to -\infty} (2x + 1) = -\infty$

So, for this case, the range is $(-\infty, 3)$.

Case 2: $x \geq 1$

• The function is constant: $h(x) = 3$.
• Therefore, for $x \geq 1$, the output value is exactly $3$.

4. Verify and Summarize

Combining both cases, we get:

• From the first case, the function yields values from $-\infty$ to just below $3$ (i.e., $(-\infty, 3)$).
• From the second case, at $x = 1$ and beyond, the function outputs exactly $3$.

Thus, the total range of $h(x)$ is: $(-\infty, 3) \cup \{3\} = (-\infty, 3]$

Final Answer

The range of $h(x)$ is $(-\infty, 3]$. Thus, the correct answer is: A. $(-\infty, 3]$