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Sagot :
To find the inverse of the function [tex]\( f(x) = x^2 + 2x \)[/tex] where [tex]\( x \geq -1 \)[/tex], follow these steps:
1. Express the function [tex]\( f(x) \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = x^2 + 2x \][/tex]
2. Rewrite the equation to solve for [tex]\( x \)[/tex]:
- Move all terms to one side to set up a quadratic equation:
[tex]\[ x^2 + 2x - y = 0 \][/tex]
- This is now in the standard form of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex].
3. Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] to solve for [tex]\( x \)[/tex]:
- Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -y \)[/tex]:
[tex]\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-y)}}{2 \cdot 1} \][/tex]
- Simplify inside the square root:
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 4y}}{2} \][/tex]
- Factor out the common term in the square root:
[tex]\[ x = \frac{-2 \pm \sqrt{4(1 + y)}}{2} \][/tex]
- Simplify further:
[tex]\[ x = \frac{-2 \pm 2\sqrt{1 + y}}{2} \][/tex]
- Divide each term inside the fraction by 2:
[tex]\[ x = -1 \pm \sqrt{1 + y} \][/tex]
4. Determine the appropriate branch of the solution:
- Since [tex]\( x \geq -1 \)[/tex], select the branch that satisfies this condition. The solution [tex]\( x = -1 - \sqrt{1 + y} \)[/tex] would always be less than or equal to -1, which does not fit our domain restriction [tex]\( x \geq -1 \)[/tex].
- Therefore, the appropriate solution is:
[tex]\[ x = -1 + \sqrt{1 + y} \][/tex]
5. Express the inverse function:
- Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to write the inverse function:
[tex]\[ x = -1 + \sqrt{1 + y} \implies y = -1 + \sqrt{1 + x} \][/tex]
Hence, the formula for the inverse of the function [tex]\( f(x) = x^2 + 2x \)[/tex] where [tex]\( x \geq -1 \)[/tex] is:
[tex]\[ f^{-1}(y) = -1 + \sqrt{1 + y} \][/tex]
This provides us with the required inverse function.
1. Express the function [tex]\( f(x) \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[ y = x^2 + 2x \][/tex]
2. Rewrite the equation to solve for [tex]\( x \)[/tex]:
- Move all terms to one side to set up a quadratic equation:
[tex]\[ x^2 + 2x - y = 0 \][/tex]
- This is now in the standard form of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex].
3. Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] to solve for [tex]\( x \)[/tex]:
- Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -y \)[/tex]:
[tex]\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-y)}}{2 \cdot 1} \][/tex]
- Simplify inside the square root:
[tex]\[ x = \frac{-2 \pm \sqrt{4 + 4y}}{2} \][/tex]
- Factor out the common term in the square root:
[tex]\[ x = \frac{-2 \pm \sqrt{4(1 + y)}}{2} \][/tex]
- Simplify further:
[tex]\[ x = \frac{-2 \pm 2\sqrt{1 + y}}{2} \][/tex]
- Divide each term inside the fraction by 2:
[tex]\[ x = -1 \pm \sqrt{1 + y} \][/tex]
4. Determine the appropriate branch of the solution:
- Since [tex]\( x \geq -1 \)[/tex], select the branch that satisfies this condition. The solution [tex]\( x = -1 - \sqrt{1 + y} \)[/tex] would always be less than or equal to -1, which does not fit our domain restriction [tex]\( x \geq -1 \)[/tex].
- Therefore, the appropriate solution is:
[tex]\[ x = -1 + \sqrt{1 + y} \][/tex]
5. Express the inverse function:
- Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to write the inverse function:
[tex]\[ x = -1 + \sqrt{1 + y} \implies y = -1 + \sqrt{1 + x} \][/tex]
Hence, the formula for the inverse of the function [tex]\( f(x) = x^2 + 2x \)[/tex] where [tex]\( x \geq -1 \)[/tex] is:
[tex]\[ f^{-1}(y) = -1 + \sqrt{1 + y} \][/tex]
This provides us with the required inverse function.
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