Answer :
To determine which of the functions \( f(x) = x^2 \), \( g(x) = 2x + 1 \), and \( h(x) = \sqrt{x} \) are one-to-one, we need to analyze each function and determine if each input \( x \) maps to a unique output \( y \).
### Function \( f(x) = x^2 \)
1. Recall that a function is one-to-one if different inputs map to different outputs.
2. Let’s assume \( f(x_1) = f(x_2) \):
[tex]\[ x_1^2 = x_2^2 \][/tex]
3. This implies:
[tex]\[ x_1 = x_2 \quad \text{or} \quad x_1 = -x_2 \][/tex]
This shows that different inputs (e.g., \( x_1 = 2 \) and \( x_2 = -2 \)) can map to the same output (e.g., \( 4 \)).
4. Therefore, \( f(x) = x^2 \) is not one-to-one.
### Function \( g(x) = 2x + 1 \)
1. Assume \( g(x_1) = g(x_2) \):
[tex]\[ 2x_1 + 1 = 2x_2 + 1 \][/tex]
2. Subtracting 1 from both sides:
[tex]\[ 2x_1 = 2x_2 \][/tex]
3. Dividing both sides by 2:
[tex]\[ x_1 = x_2 \][/tex]
This shows that different inputs \( x_1 \) and \( x_2 \) must be the same if the outputs are the same.
4. Therefore, \( g(x) = 2x + 1 \) is one-to-one.
### Function \( h(x) = \sqrt{x} \)
1. Assume \( h(x_1) = h(x_2) \):
[tex]\[ \sqrt{x_1} = \sqrt{x_2} \][/tex]
2. Squaring both sides:
[tex]\[ x_1 = x_2 \][/tex]
This shows that different inputs \( x_1 \) and \( x_2 \) must be the same if the outputs are the same.
3. Therefore, \( h(x) = \sqrt{x} \) is one-to-one on its domain (non-negative real numbers).
Based on this analysis:
- \( f(x) = x^2 \) is not one-to-one (because it maps both \( x \) and \( -x \) to the same value).
- \( g(x) = 2x + 1 \) is one-to-one.
- \( h(x) = \sqrt{x} \) is one-to-one.
Therefore, the correct answer is B: [tex]\( g \)[/tex] only.
### Function \( f(x) = x^2 \)
1. Recall that a function is one-to-one if different inputs map to different outputs.
2. Let’s assume \( f(x_1) = f(x_2) \):
[tex]\[ x_1^2 = x_2^2 \][/tex]
3. This implies:
[tex]\[ x_1 = x_2 \quad \text{or} \quad x_1 = -x_2 \][/tex]
This shows that different inputs (e.g., \( x_1 = 2 \) and \( x_2 = -2 \)) can map to the same output (e.g., \( 4 \)).
4. Therefore, \( f(x) = x^2 \) is not one-to-one.
### Function \( g(x) = 2x + 1 \)
1. Assume \( g(x_1) = g(x_2) \):
[tex]\[ 2x_1 + 1 = 2x_2 + 1 \][/tex]
2. Subtracting 1 from both sides:
[tex]\[ 2x_1 = 2x_2 \][/tex]
3. Dividing both sides by 2:
[tex]\[ x_1 = x_2 \][/tex]
This shows that different inputs \( x_1 \) and \( x_2 \) must be the same if the outputs are the same.
4. Therefore, \( g(x) = 2x + 1 \) is one-to-one.
### Function \( h(x) = \sqrt{x} \)
1. Assume \( h(x_1) = h(x_2) \):
[tex]\[ \sqrt{x_1} = \sqrt{x_2} \][/tex]
2. Squaring both sides:
[tex]\[ x_1 = x_2 \][/tex]
This shows that different inputs \( x_1 \) and \( x_2 \) must be the same if the outputs are the same.
3. Therefore, \( h(x) = \sqrt{x} \) is one-to-one on its domain (non-negative real numbers).
Based on this analysis:
- \( f(x) = x^2 \) is not one-to-one (because it maps both \( x \) and \( -x \) to the same value).
- \( g(x) = 2x + 1 \) is one-to-one.
- \( h(x) = \sqrt{x} \) is one-to-one.
Therefore, the correct answer is B: [tex]\( g \)[/tex] only.
