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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]

Question

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Solution

1. Break Down the Problem

We need to find the range of the piecewise function h(x) h(x) , which is defined as follows: h(x)={2x+1if x<13if x1 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 x :

  1. For x<1 x < 1 : Evaluate h(x)=2x+1 h(x) = 2x + 1 .
  2. For x1 x \geq 1 : Evaluate h(x)=3 h(x) = 3 .

3. Analysis and Detail

Case 1: x<1 x < 1

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

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

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

Case 2: x1 x \geq 1

  • The function is constant: h(x)=3 h(x) = 3 .
  • Therefore, for x1 x \geq 1 , the output value is exactly 3 3 .

4. Verify and Summarize

Combining both cases, we get:

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

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

Final Answer

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

This problem has been solved

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