+79 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!!
+2446 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.
2. Replace 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. | ||
| 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.
