# Math Help - inverse function question

1. ## inverse function question

the function is f(x) = 6x + x^2
I need to find the largest restricted domain for which f has an inverse
and to explain the answer carefully..
2nd part is to find the inverse using the original function and the domain i found in part 1 and state its domain and range.
i'm afraid it has be a bit stumped..
help much appreciated x

2. Originally Posted by synnr
the function is f(x) = 6x + x^2
I need to find the largest restricted domain for which f has an inverse
and to explain the answer carefully..
we need the quadratic to be one-to-one in order for its inverse to be a function. we can accomplish this by splitting the quadratic into two parts divided by the vertex, and then picking one of those parts.

so, if, for example, the x coordinate of the vertex is $x_v$, then the largest restricted domain for which f has an inverse function is $(- \infty, x_v]$ or $[ x_v, \infty)$

2nd part is to find the inverse using the original function and the domain i found in part 1 and state its domain and range.
we find the inverse function by switching x and y and solving for y

that is, take $y = 6x + x^2$

then switch x and y to obtain

$x = 6y + y^2$

now solve for y

the domain and range of your inverse function depends on what interval you chose for the first question, but i will say this, the domain of your original function is the range of your inverse function, and the range of your inverse function is the domain of your original function

do you know how to find the domain and range?

can you continue?

3. great! that really helps, thanks.
I do understand how to find inverse functions its just this one really has me spinning..
i've done a whole page of attempted working out but i just seem to have a block where squareroots are involved.
the closest i got was.. f^-1 (x) = (sqrt x/6) / 2 but that is so ugly and so wrong..!
thanks again

4. Originally Posted by synnr
great! that really helps, thanks.
I do understand how to find inverse functions its just this one really has me spinning..
i've done a whole page of attempted working out but i just seem to have a block where squareroots are involved.
the closest i got was.. f^-1 (x) = (sqrt x/6) / 2 but that is so ugly and so wrong..!
thanks again
think of it this way:

we have: $x = 6y + y^2$

$\Rightarrow y^2 + 6y - x = 0$

now, it may take a while for you to see it, but this is, in fact, a quadratic equation in y. with a = 1, b = 6 and c = -x. we can use the quadratic formula to solve for y

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

$\Rightarrow y = - \frac 62 \pm \frac {\sqrt{36 + 4x}}2$

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

now whether you take the + square root, or the - square root, depends on which of the two intervals you chose in the first part