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Sagot :
To find the inverse of the function [tex]\(f(x) = 2x + 1\)[/tex], follow these steps:
1. Start with the given function:
[tex]\[ f(x) = 2x + 1 \][/tex]
2. Replace [tex]\(f(x)\)[/tex] with [tex]\(y\)[/tex] for easier manipulation:
[tex]\[ y = 2x + 1 \][/tex]
3. Solve for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]. First, isolate the term with [tex]\(x\)[/tex]:
[tex]\[ y - 1 = 2x \][/tex]
4. Next, solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{y - 1}{2} \][/tex]
5. Now replace [tex]\(y\)[/tex] with [tex]\(x\)[/tex] to write the inverse function:
[tex]\[ h(x) = \frac{x - 1}{2} \][/tex]
6. Simplify the expression:
[tex]\[ h(x) = \frac{1}{2}x - \frac{1}{2} \][/tex]
This matches the form given in one of the options. Therefore, the inverse function is:
[tex]\[ h(x) = \frac{1}{2}x - \frac{1}{2} \][/tex]
Given the choices:
- [tex]\( h(x) = \frac{1}{2} x - \frac{1}{2} \)[/tex]
- [tex]\( h(x) = \frac{1}{2} x + \frac{1}{2} \)[/tex]
- [tex]\( h(x) = \frac{1}{2} x - 2 \)[/tex]
- [tex]\( h(x) = \frac{1}{2} x + 2 \)[/tex]
The correct answer is:
[tex]\[ h(x) = \frac{1}{2}x - \frac{1}{2} \][/tex]
1. Start with the given function:
[tex]\[ f(x) = 2x + 1 \][/tex]
2. Replace [tex]\(f(x)\)[/tex] with [tex]\(y\)[/tex] for easier manipulation:
[tex]\[ y = 2x + 1 \][/tex]
3. Solve for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]. First, isolate the term with [tex]\(x\)[/tex]:
[tex]\[ y - 1 = 2x \][/tex]
4. Next, solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{y - 1}{2} \][/tex]
5. Now replace [tex]\(y\)[/tex] with [tex]\(x\)[/tex] to write the inverse function:
[tex]\[ h(x) = \frac{x - 1}{2} \][/tex]
6. Simplify the expression:
[tex]\[ h(x) = \frac{1}{2}x - \frac{1}{2} \][/tex]
This matches the form given in one of the options. Therefore, the inverse function is:
[tex]\[ h(x) = \frac{1}{2}x - \frac{1}{2} \][/tex]
Given the choices:
- [tex]\( h(x) = \frac{1}{2} x - \frac{1}{2} \)[/tex]
- [tex]\( h(x) = \frac{1}{2} x + \frac{1}{2} \)[/tex]
- [tex]\( h(x) = \frac{1}{2} x - 2 \)[/tex]
- [tex]\( h(x) = \frac{1}{2} x + 2 \)[/tex]
The correct answer is:
[tex]\[ h(x) = \frac{1}{2}x - \frac{1}{2} \][/tex]
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