# Thread: help with sign graphs and inverse polynomial functions

1. ## help with sign graphs and inverse polynomial functions

I have some questions that I have the answers for but no idea how to reach those answers.

1)
x^3 - 6x^2 +5x + 6 < 0 (use sign graph)
the < is supposed to be "less than or equal to".
the answer I'm given is: (-inf, 2-sqrt7] U [2, 2+sqrt7]
I've never learned what that way of writing the answer means though.

2) find f-1(x) (it's supposed to be find the inverse)
f(x) = -(x-2)^2, x < 2
the < is supposed to be "less than or equal to".
the answer I'm given is: f-1(x) = 2-sqrt(-x)

3) find f-1(x) (it's supposed to be find the inverse)
f(x) = (2x+1) / (x+3), x cannot equal 3.
the answer I'm given is: f-1(x) = (1-3x) / (x-2)

-inf means negative infinity, sqrt means square root of

2. Originally Posted by DarthBlood
I have some questions that I have the answers for but no idea how to reach those answers.

1)
x^3 - 6x^2 +5x + 6 < 0 (use sign graph)
the < is supposed to be "less than or equal to".
the answer I'm given is: (-inf, 2-sqrt7] U [2, 2+sqrt7]
I've never learned what that way of writing the answer means though.[snip]
The cubic factorises as $\displaystyle (x^2 - 4x - 3)(x - 2)$. So draw the graph of $\displaystyle y = (x^2 - 4x - 3)(x - 2)$ (be sure to find the three x-intercepts). The solution to the given inequality is the set of values of x such that y < 0.

Originally Posted by DarthBlood
[snip]
2) find f-1(x) (it's supposed to be find the inverse)
f(x) = -(x-2)^2, x < 2
the < is supposed to be "less than or equal to".
the answer I'm given is: f-1(x) = 2-sqrt(-x)
[snip]
$\displaystyle y = f^{-1}(x)$ where $\displaystyle x = -(y - 2)^2$ and y < 2. Solve for y.

Originally Posted by DarthBlood
[snip]
3) find f-1(x) (it's supposed to be find the inverse)
f(x) = (2x+1) / (x+3), x cannot equal 3.
the answer I'm given is: f-1(x) = (1-3x) / (x-2)

-inf means negative infinity, sqrt means square root of

$\displaystyle y = f^{-1}(x)$ where $\displaystyle x = \frac{1 - 3y}{y - 2}$. Solve for y by making it the subject:

$\displaystyle x = \frac{1 - 3y}{y - 2} \Rightarrow x(y - 2) = 1 - 3y \Rightarrow xy - 2x = 1 - 3y$ etc.

3. i understand the first 2, but i still don't get the 3rd one.

when you change the x and y, you changed using the answer. shouldn't i be changing with the original equation?

so that the result is:

$\displaystyle x = \frac{2y+1}{y+3}$

not,

$\displaystyle x = \frac{1 - 3y}{y - 2}$

4. Originally Posted by DarthBlood
i understand the first 2, but i still don't get the 3rd one.

when you change the x and y, you changed using the answer. shouldn't i be changing with the original equation?

so that the result is:

$\displaystyle x = \frac{2y+1}{y+3}$

not,

$\displaystyle x = \frac{1 - 3y}{y - 2}$$\displaystyle$
You're right. I misread f^-1 as the given function. The idea is exactly the same though.

5. ah i got it! thanks a lot!