Question

For the function f(x) = (8-2x)2 ,find f-1 . Determine whether f-1 is a function.

Answer 1

\[\Large y = (8-2x)^2\]
solve for x

start by taking square roots of both sides

Answer 2

sqrts of both sides?

Answer 3

\(\bf \sqrt{y}=\sqrt{(8-2x)^2}\leftarrow that\)

Answer 4

And what do i do with that?

Answer 5

Square root cancels out a square

Answer 6

i dont get it ?

Answer 7

\[\large \sqrt {x^2} = x\]
\[\large \sqrt{(x-6)^2} = x-6\]
etc

Answer 8

oh okay so it would be the same thing is was before it was sqrt with out the 2?

Answer 9

Er, not sure what you mean, just write it.

Answer 10

y = (8-2x)

Answer 11

Sqrt y is still sqrt y.

Answer 12

oh so its sqrt y= (8-2x) ?

Answer 13

\(\bf \sqrt{y}=\sqrt{(8-2x)^2}
\\ \quad \\
\sqrt[{\color{red}{ 2}}]{x^{\color{red}{ 2}}} \iff x\qquad thus
\\ \quad \\
\sqrt{y}=\sqrt[{\color{red}{ 2}}]{(8-2x)^{\color{red}{ 2}}}\implies \sqrt{y}=8-2x\)

Answer 14

Now subtract 8 from both sides

then divide by -2

Answer 15

sqrty -8= -2x?

Answer 16

Yes

Answer 17

now what?

Answer 18

See my last post, it's there

Answer 19

Which one?

Answer 20

Now subtract 8 from both sides

then divide by -2

Answer 21

x=sqrt y-8/ -2 ?

Answer 22

you forgot parentheses :P

\[\large x =\frac{ ( \sqrt y -8 ) }{-2 }\]

Answer 23

Simplify, by multiplying the top and bottom of the fraction by -1:
\[\large x =\frac{ 8- \sqrt y }{2 }\]now just swap x and y.

Answer 24

y= 8- sqrt x/2 ?

Answer 25

Its not a function right?

Answer 26

It's not because we forgot to put the plus or minus in, which you have to do when taking square roots

and you forgot parentheses again :P

Answer 27

\[\large x =\frac{ 8\pm \sqrt y }{2 }\]

Answer 28

Okay :) thanks :)