Question

i need to find a formula for f^-1(x) when f(x)=x^3

Answer 1

f(x) = y = x³

Now switch x and y to each other.

x = y³ This is your inverse function.

Now isolate y.

Answer 2

Still not clear?

Answer 3

@SugarRainbow

Answer 4

sorry i really quick

Answer 5

i was eating

Answer 6

y=[\[Y=\sqrt[3]{X}\]x^(1/3)]

Answer 7

Yup, this is your inverse function.

Answer 8

i actually got Y=X−−√3

Answer 9

i mean i got what kevinbourne thied got except for the x^(1/3) how did he get that?

Answer 10

IT means same thing, just another way to write \(\Large \sqrt[3]{x}\)

Answer 11

is ti better to write it the other way or does it not really matter?

Answer 12

I don't think it matters.

Answer 13

okay so for this next question i need to find the inverse of f(x)=x^3+5
i got x^(1/3)-5
correct?

Answer 14

x=y3+5
y3=-x-5
y=[(-x)^(1/3)]-5

Answer 15

I believe it's (x-5)^(1/3)

Answer 16

what why -x? more explanation please from kevinbournethird
and geerky tell me how you got yours?

Answer 17

I put my steps in my answer.

Answer 18

i know but how and why did you get -x?

Answer 19

Do the same thing: switch x and y to each other.
x³ + 5 = y
y³ + 5 = x
y³ = x - 5
y = ³√(x - 5)

Answer 20

but hold on let me check somethin

Answer 21

wait that is what i got except i didn't group the x-5 together in parentheses why u do that?

Answer 22

So we know that '-5' is part of the cube root.

You also can write it like that: \(\large y = \sqrt[3]{x-5}\)

Answer 23

x^(1/3)-5 looks like \(\sqrt[3]{x} - 5\)

Answer 24

yeah but see i was thinkin of it like you find the 3root of x and then subtract five or is it a little backwards since it IS an inverse and ervrythings backwards?

Answer 25

well, inverse is just a function that was flipped over line y=x so we switch x and y to each other then isolate y.

Answer 26

okay then so the next question is
so would it be like the previous one and be
x^3-5?