I need to find the inverse function of f(x)=sqrt(x^2+2x), where x>o
This is as far as I got:
x=sqrt(y^2+2y)
How in the world am I going to isolate for y? Should I complete the square underneath the square root symbol, and then use the quadratic equation? Does that work? :/
y = sqrt(x^2 + 2x)
square both sides
y^2 = x^2 + 2x
x^2 + 2x = y^2
complete the square
x^2 + 2x + 1 = y^2 + 1
(x + 1)^2 = y^2 + 1
take the square root of both sides
+ x + 1 = (sqrt(y^2 + 1))
- x - 1 = (sqrt(y^2 + 1))
x = -1 + - (sqrt(y^2 + 1))
interchange x and y
y = -1 +- (sqrt(x^2 + 1))
To find the inverse function of f(x) = √(x^2+2x) where x > 0, you are on the right track.
x = √(y^2+2y)
To isolate y, square both sides of the equation:
x^2 = y^2 + 2y
Rearrange the equation:
y^2 + 2y = x^2
Now complete the square on the left side:
(y^2 + 2y + 1) - 1 = x^2
(y + 1)^2 = x^2 + 1
Take the square root of both sides, noting that y > 0:
y + 1 = √(x^2 + 1)
Solve for y:
y = √(x^2 + 1) - 1
So, the inverse function is:
f^(-1)(x) = √(x^2 + 1) - 1
To find the inverse of a function, we need to isolate the input variable (usually expressed as y) on one side of the equation.
Given the equation: x = √(y^2 + 2y)
To isolate for y, one approach is to square both sides of the equation:
x^2 = y^2 + 2y
Next, rearrange the equation to get all terms on one side and set it equal to zero:
y^2 + 2y - x^2 = 0
Now, we have a quadratic equation. To solve for y, we can try completing the square or use the quadratic formula.
Using the quadratic formula, the equation becomes:
y = (-2 ± √(2^2 - 4(1)(-x^2))) / (2(1))
Simplifying further:
y = (-2 ± √(4 + 4x^2)) / 2
y = (-2 ± 2√(1 + x^2)) / 2
y = -1 ± √(1 + x^2)
Since x > 0 in the given function, we can discard the negative square root since it will yield a negative value for y.
Therefore, the inverse function is:
f^(-1)(x) = √(1 + x^2) - 1
Note that the inverse function exists only for x > 0 because the original function's domain restricts x to be greater than 0.