How do you find the inverse of #f(x)=x^2-6x#?

Question

Elijah Allen · Accepted Answer

The function is not one-to-one, so it does not have an inverse function.

The inverse relation may be found by solving #x = y^2-6y# for #y# using either Completing the Square or the Quadratics Formula: #y^2-6y-x=0# We have #a=1#, b=-6# and #c=-x# #y = (-b +- sqrt(b^2-4ac))/2a# #y = (-(-6) +- sqrt((-6)^2-4(1)(-x)))/(2(1))# # y = (6+-sqrt(36+4x))/2# # y = (6+-sqrt(4(9+4x)))/2# # y = (6+- 2sqrt(9+4x))/2# #y = 3 +- sqrt(9+x)# As we can see #y# is not a function of #x#. That is: the inverse relation is not a function.

Elijah Allen · Answer

This function does not have an inverse.

We have that #f(x)=(x^2-6x+9)-9=(x-3)^2-9# hence #f(0)=f(6)# this function is not #1-1# This function #f:R->R# does not have an inverse.

James Allen · Answer

undefined To find the inverse of the function $f(x) = x^2 - 6x$, you follow these steps:

1. First, replace $f(x)$ with $y$: $y = x^2 - 6x$.

2. Swap $x$ and $y$: $x = y^2 - 6y$.

3. Solve this equation for $y$. This involves rearranging the equation to follow the standard form of a quadratic equation and then applying the quadratic formula. The standard form is $ay^2 + by + c = 0$, so you have $y^2 - 6y - x = 0$.

Applying the quadratic formula, where $a = 1$, $b = -6$, and $c = -x$, we get:
$$y = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-x)}}{2 \cdot 1}$$

$$y = \frac{6 \pm \sqrt{36 + 4x}}{2}$$

$$y = 3 \pm \sqrt{9 + x}$$

Since $f(x) = x^2 - 6x$ is a parabola opening upwards with a vertex at $(3, -9)$, the function is not one-to-one over its entire domain (it does not pass the Horizontal Line Test). Therefore, to have an inverse, we must restrict the domain. Assuming the parabola is restricted to $x \geq 3$ (right side of the parabola), the correct sign in the inverse function would be the plus sign to ensure the output is in the restricted domain.

So, the inverse function, $f^{-1}(x)$, for $x \geq 3$ is:
$$f^{-1}(x) = 3 + \sqrt{9 + x}$$