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# Do these inverse functions pass the vertical line test?

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Basically, i have two functions that i needed to find inverses of.  They are -
1. f(x)=2x2 +2x-1
2. f(x)=-4.9(x+3)2+45.8

Thanks to hecticar and CPhill, the inverses were found. They can be seen in the attachment below. Now I need to prove whether the inverses are actually functions or not. I plan to do it with the vertical line test. I understand that you can not enter ± on Desmos and show both values simultaneously. Should I input the two values as seperate functions? How exactly do i go about correctly entering these complicated inverse functions into Desmos?I have no prior experience with functions that have two values, a positive and negative one. Can someone please clarify? Would be a life saver for sure!!

Sep 16, 2017

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Finding the inverse of an equations requires a few steps. I will use the orginal function $$f(x)=2x^2+2x-1$$

1. Change $$f(x)$$ to $$y$$.

This means that $$f(x)=2x^2+2x-1$$ changes to $$y=2x^2+2x-1$$. That is all that this step entails.

2Replace all instances of y with x and all instances of y with x

This step is relatively simple, too.

$$y=2x^2+2x-1$$ changes to $$x=2y^2+2y-1$$. Now, this step is done.

3. Solve for y

Solving for y means to isolate it. Since the quadratic formula is only a valid option when solving for the root of the equation, we cannot use that method with multiple variables. It appears as if completing the square is the only option.

$$2y^2+2y-1=x$$We need to get the "c" term on the opposite side of the equation for completing the square. Therefore, add 1 to both sides.
$$2y^2+2y=x+1$$Since the a-term must be one in order for completing the square to work, we must divide the entire equation by 2.
$$y^2+y=\frac{x+1}{2}$$This is the trickiest bit. We need to make the lefthand side a perfect square. To do this, add $$\left(\frac{b}{2}\right)^2$$ where b is the coefficient of the linear term. You must add it to both sides because whatever you do to one side, you must do to the other.
$$y^2+y+\left(\frac{b}{2}\right)^2=\frac{x+1}{2}+\left(\frac{b}{2}\right)^2$$Replace b with the coefficient of the linear term, 1.
$$y^2+y+\left(\frac{1}{2}\right)^2=\frac{x+1}{2}+\left(\frac{1}{2}\right)^2$$Simplify both sides.
$$y^2+y+\frac{1}{4}=\frac{x+1}{2}+\frac{1}{4}$$The left hand side is now a perfect-square, so transform it into one.
$$\left(y+\frac{1}{2}\right)^2=\frac{x+1}{2}+\frac{1}{4}$$Before taking the square root of both sides, we should add the fractions together.
$$\frac{x+1}{2}+\frac{1}{4}=\frac{2x+2}{4}+\frac{1}{4}=\frac{2x+3}{4}$$Reinsert this back into the equation.
$$\left(y+\frac{1}{2}\right)^2=\frac{2x+3}{4}$$Take the square root of both sides.
$$y+\frac{1}{2}=\pm\sqrt{\frac{2x+3}{4}}$$Distribute the square root to both the numerator and denominator. Remember that taking the square root results in the positive and negative answer.
$$y+\frac{1}{2}=\pm\frac{\sqrt{2x+3}}{2}$$Subtract 1/2 from both sides.
$$y=\pm\frac{\sqrt{2x+3}-1}{2}$$Break them up into separate solutions.
 $$y=\frac{\sqrt{2x+3}-1}{2}$$ $$y=\frac{-\sqrt{2x+3}+1}{2}$$

We can convert this into function notation, if you want.
$$f^{-1}(x)=\pm\frac{\sqrt{2x+3}-1}{2}$$

I will get to graphing later.

The next function is $$f(x)=-4.9(t+3)^2+45.8$$

Let's do the same steps again. Change f(x) to y and flip flop all "t's" and "y's."

$$t=-4.9(y+3)^2+45.8$$

Last time, I solved the equation, but I won't do that with this function because it appears as if you did it correctly.

Ok, inputting it into Desmos is easier than you might think.

Equation 1: $$y=2x^2+2x-1$$

Equation 1 Inverse: $$x=2y^2+2y-1$$

This is the inverse because every x value is now the y-value. We can do the same for equation 2/

Equation 2: $$y=-4.9(t+3)^2+45.8$$

Equation 2 Inverse: $$t=-4.9(y+3)^2+45.8$$

Now, just put in both the "Equation 1 Inverse" and "Equation 2 Inverse" into Desmos, and Desmos figure out the rest. Click here to see the graph is Desmos. Spoiler Alert: both equations fail the vertical line test.

Sep 16, 2017
edited by TheXSquaredFactor  Sep 16, 2017
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Reflection on the line y = x

Sep 16, 2017