Question

Hi, I need help finding the inverse of the function f(x) = x^2 + 2. I've gotten to x = y^2 + 2, and I don't know how to solve for y here. Please explain the process in steps, and I'll do the work myself?

Answer 1

start by subtracting 2 from both sides
that's isolate y^2

and then taking square root on both side, isolates y :)

Answer 2

So would that be \[\sqrt{x - 2} = \sqrt{y ^{2}}\]

Answer 3

yeah and considering positive root,
\(\sqrt {y^2} = y\)

Answer 4

so your inverse function is just \(\Large \sqrt{x-2}\)
thats it!

Answer 5

Would I write it as
y = \[\sqrt{x - 2}\]
or...?

Answer 6

f^-1(x) = sqrt x - 2?

Answer 7

the second one :)

Answer 8

\(\Large f^{-1}(x) = \sqrt {x-2}\)

Answer 9

Thanks so much! Can you help me with another one?

Answer 10

While simplifying some math work, Peter wrote on his paper that x^3 * x^3 * x^3 * x^3 = x^3 + the power of 3 + the power of 3 + the power of 3. Does this show the final answer? Would Peter's work be the same if he were to simplify x^3 + x^3 + x^3 + x^3?

I wrote no, because you have to multiply exponents, not add them together, right? It wouldn't be x^12, it would be x^3 four times.

Answer 11

x^3*x^3*x^3*x^3=x^(3+3+3+3)=x^(12)
is right
and you did good saying x^3+x^3+x^3+x^3=4x^3

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and that first one
y=x^2+2 is non a one-to-one function so it doesn't have an inverse function.

if you restrict the domain of y=x^2+2 from x=0 to infinity then the inverse is y=sqrt(x-2)

Answer 12

With the domain restriction part, that's what we did :) I just didn't know that's what it was called, lol.

Answer 13

Thanks so much!!