Question

Let f(x) = x^2 – 16. Find f^–1(x).

Answer 1

Let f(x) = y

So,

\[y = x^2 - 16\]

\[x^2 = y + 16\]

\[x = \sqrt{y+16}\]

Just swap x and y now:

\[y = \sqrt{x + 16}\]

This is \(f^{-1}(x)\)..

Answer 2

@waterineyes...why not ! \[-\sqrt{y+16}\]

Answer 3

Yes it can be but I here only consider the positive value of root..

Answer 4

\[\pm 4\sqrt{}x\]
\[\pm \sqrt{+ 16} \]
\[x ^{2} \div 16\]
\[1\div x^{2} - 16\]

Answer 5

one of those have to be the answers

Answer 6

then, they mean 1/f(x)...it's number 4 !

Answer 7

the last question ?

Answer 8

go on if you have more questions :)

Answer 9

lol so the answer is the fourth . correct?

Answer 10

yeaaah...\[1/(x^2-16)\]

Answer 11

ohh ok thank youu (:

Answer 12

yw ! anytime !