Question

f(x)= 2x^2 - 3

Answer 1

Find the zeroes?

Answer 2

inverse function of f(x)= 2x^2 - 3

Answer 3

Okay, first input y for f(x).

\(f(x) = 2x^2 - 3\)
\(y = 2x^2 - 3\)

Answer 4

apply the inverse operations in the opposite order

Answer 5

Then switch x and y:

\(x = 2y^2-3\)

Answer 6

Then solve for \(y\).

Answer 7

what is the answer? :)

Answer 8

\(x = 2y^2 - 3\)

\(\Large\frac{x}{2} \) \(= y^2 - 3\)

Answer 9

I will not tell you the answer, you will find it out on your own.

Answer 10

Am I right so far? @skullpatrol

Answer 11

Thank You :) Very much

Answer 12

nope, first add 3 to both sides

Answer 13

Oh.

Answer 14

How to find?

Answer 15

\(x=2y^2−3\)
\(x + 3 = 2y^2\)

Answer 16

Do we divide by 2 or find the exponent next? @skullpatrol

Answer 17

Or square the other side*

Answer 18

divide by 2

Answer 19

Okay, can you do that @jhayferrer

Answer 20

$$\Huge \frac{x+3}{2}=y^2$$

Answer 21

oh that's the answer men :)

Answer 22

Argh..

Hold on.

\[y = \sqrt{\frac{x+3}{2}}\]

Answer 23

and then?

Answer 24

That's your final answer, do you see how we did it?

Answer 25

no :(

Answer 26

you need the plus or minus sign

Answer 27

Well you change it to:

\(f^{-1}(x) = \sqrt{\Large\frac{x+3}{2}}\)

Answer 28

Where? @skullpatrol

Answer 29

in front of the square root

Answer 30

Oh.

Answer 31

i find thank you very much

Answer 32

How about inverse function of f(x)=x

Answer 33

Okay, well @jhayferrer

First we changed \('f(x)'\) to \(y\).

\(y= 2x^2 - 3\)

Then we switched the \(x\) and \(y\) variables.

\(x = 2y^2 - 3\)

Now we're trying to get the \(y\) variable by itself. So we have to add the \(3\).

\(x + 3 = 2y^2\)

Then we divide the \(2\) because it's being multiplied by the \(y\).

\(\Large\frac{x+3}{2}\)\( = y^2\)

Now to get rid of the exponent \('2'\), we have to square the other side:

\(y = \sqrt{\Large\frac{x+3}{2}}\)

Answer 34

Then:

\(f^{-1}(x) = \pm\sqrt{\Large\frac{x+3}{2}}\)

Answer 35

\(f(x)=x\)

Change the \(f(x)\)to \(y\), then switch the \(x\) and \(y\).

Answer 36

Now what do we have? @jhayferrer