INVERSE function

[vocal panther]

To be clear, this does not have an inverse for all real numbers. Do you mean to find the inverse for x >= 3?

[grave hawk]

Idk my teacher only gave this and he said that we need to find the inverse here & needed the complete solution

[vocal panther]

Ah, well the teacher may be leaving off details. Anyway, what have you tried so far?

[grave hawk]

Here hahahaha

[vocal panther]

Ok, good, you swapped out the x and the y. Next you should solve for y.

[grave hawk]

How should I solve the Y? If I have ²?

[vocal panther]

An important idea: Cancellation. I'll show you. First, we move the -2 to the other side and get $x+2 = (y-3)^2$, right?

[sonic frost]

addem

[vocal panther]

Next we ask "what cancels a square?" The answer is: A square-root.

So we use the square-root on both sides. Makes sense so far?

[grave hawk]

Yep

Wait I'll show u

[vocal panther]

Cool, so if we use the square root we get $\sqrt{x+2}=\sqrt{(y-3)^2}$ but because square-roots cancel squares, then this is actually the same thing as $\sqrt{x+2}=y-3$. Note that this solution is only valid when $x\ge -2$.

[sonic frost]

addem

[grave hawk]

???

So it would be x+2 = y-3?

After cancelation?

[vocal panther]

No, it's $\sqrt{x+2} + 3 = y$.

[sonic frost]

addem

[grave hawk]

Ohh so I need to swapped the 3?

[vocal panther]

Well it just gets moved to the other side. After cancellation it's $\sqrt{x+2}=y-3$ but you want to keep solving for $y$ so you add the 3 to the other side and that gets the answer.

[sonic frost]

addem

[grave hawk]

Like this? So after that

Im goin to move -3 so it will become +3

I'm just curious how to disappear the ()²?

[grave hawk]

I must first do the square root I think?

Idk what should I do first huhuhu

My teacher ask for complete solution

[vocal panther]

So here's the whole solution: $f(x)=(x-3)^2 -2$. Then $y = (x-3)^2-2$. Then for the inverse function we use $x = (y-3)^2-2$. Then $x+2=(y-3)^2$. Then $\sqrt{x+2}=\sqrt{(y-3)^2}$ then $\sqrt{x+2}=y-3$ then $\sqrt{x+2}+3=y$. So the inverse function is $f^{-1}(x) = \sqrt{x+2}+3$.

[sonic frost]

addem

[grave hawk]

Thank u so muchhhh🥹🫶

[vocal panther]

👍

[grave hawk]

I finally get it, it's hard for me because I thought that -3 do not needed to moved

[vocal panther]

Right, it's just the final step to actually get y, it just happens after the "big" step which is cancellation.

[grave hawk]

Thank u so much🫶 I need to finish it today so I will be able to upload my task on gclass because today was the deadline huhuhu

Here's my number 5 idk if I am correct here

[vocal panther]

Ok, I'll look at one more but then I need to do my own work lol

[grave hawk]

Oh okay thank u so much 🫶

[vocal panther]

Ok, it looks like at the end you kinda figured it out. I agree with $x(y^3-5) = 8$ so that's good. Now we're solving for $y$ so we move $x$ to the other side and get $y^3-5=\frac 8 x$. We still want to get things away from $y$ so we add 5, getting $y^3 = \frac 8 x + 5$.

[sonic frost]

addem

[vocal panther]

Finally, we take the cube-root to get $y$. that gets us $y = \sqrt[3]{\frac 8 x + 5}$. So the inverse function is $f^{-1}(x) = \sqrt[3]{\frac 8 x + 5}$.

[sonic frost]

addem

[grave hawk]

Ahh okay I thought that I need to isolate the dependent variable