For the life of me I can't figure this out!

[Amira88]

Hi everyone,

I am coming from a social science background and I am currently doing Grade 12 math. I am having the hardest time finding the inverse of the following function:

y = 2x^2 + 2x - 1

I inverse both variables and get to this, but beyond this point I am stuck.

(x+1)/2 = y^2 + y

 

[tkhunny]

"inverse both variables" doesn't mean anything,

The standard procedure is to EXCHANGE x and y and then re-solve for y.

You solve for y, in this case, by COMPLETING THE SQAURE. You're almost there.

(x+1)/2 + 1/4 = y^2 + y + 1/4 = (y+1/2)^2

It's close. Where did 1/4 and 1/2 come from? Of course, this is also the moment when you must decide if your inverse is a FUNCTION or not.

 

[pka]

I am coming from a social science background and I am currently doing Grade 12 math. I am having the hardest time finding the inverse of the following function: y = 2x^2 + 2x - 1

Because this function is not an injection(one-to-one) there is no inverse.

 

[JeffM]

PKA is of course correct. You can find an inverse of a function only in a domain where the function is nowhere decreasing or nowhere increasing.

With respect to your function, it is not decreasing if x is greater than or equal to -1/2. It is not increasing if x is less than or equal to -1/2. Thus, we get different inverses in different domains. Let's look at the case where x is non-negative and y is everywhere increasing as x increases.

You can complete the square or use the quadratic formula.

[MATH]f(x) = 2x^2 + 2x - 1.[/MATH]
[MATH]x = 2y^2 + 2y - 1 \implies 2y^2 + 2y - 1 - x = 0 \implies [/MATH]
[MATH]y = \dfrac{-2 \pm \sqrt{4 -4(2)(-1 - x)}}{2 * 2} = \dfrac{-1 \pm \sqrt{3 + 2x}}{2}.[/MATH]
Do we want plus or minus? Well when y is increasing, x is increasing. So we want plus.

[MATH]f^{-1}(x) = \dfrac{-1 + \sqrt{3 + 2x}}{2}.[/MATH]
[MATH]f^{-1}(f(x)) = f^{-1}(2x^2 + 2x - 1) = \dfrac{-1 + \sqrt{3 + 4x^2 + 4x - 2}}{2} = \dfrac{-1 + 2\sqrt{x^2 + x + 0.25}}{2} =\\ \dfrac{-1 + 2 \sqrt{(x + 0.5)^2}}{2} = \dfrac{-1 + 2(x + 0.5)}{2} = \dfrac{2x + 1 - 1}{2} = x \ \checkmark.[/MATH]
But is it going the right direction?

[MATH]f(0) = -1.[/MATH]
[MATH]f^{-1}(-1) = \dfrac{-1 + \sqrt{3 + 2(-1)}}{2} = \dfrac{-1 + \sqrt{1}}{2} = 0 \ \checkmark.[/MATH]
[MATH]f(1) = 3.[/MATH]
[MATH]f^{-1}(3) = \dfrac{-1 + \sqrt{3 + 2(3)}}{2} = \dfrac{-1 + \sqrt{3 + 6}}{2} = 1 \ \checkmark.[/MATH]
Looks good, but kind of ugly getting there.

The fact that a function does not have an inverse everywhere still may permit it to have an inverse in the domain that is relevant to the problem you are working on.

 

[Dr.Peterson]

A function that is not one-to-one doesn't have an inverse function; but it does have an inverse relation, which is what you are finding. It has two values in many cases.