Question

Which of the following statements have the same result? Explain each step in solving each one.

f(2) when f(x) = 3x + 2
f-1(3) when f(x) = 2 x minus 7, all over 3
2y + 14 = 4y - 2

Answer 1

you would need to calculate each thing.
For part A.
\(\large\color{black}{ f(x)=3x+2 }\) and if you wanted to find \(\large\color{blue}{ f(3) }\) you would just plug in 3 for x,
\(\large\color{black}{ f(\color{blue}{3})=3\color{blue}{(3)}+2 }\)
\(\large\color{black}{ f(\color{blue}{3})=9+2 }\)
\(\large\color{black}{ f(\color{blue}{3})=11 }\)

Answer 2

So, if you were to find \(\large\color{black}{ f(\color{blue}{2}) }\) how would you do this?

Answer 3

If(x) = 3(2) + 2
f(3) = 6 + 2
f(3) = 8

Answer 4

you mean f(2), not f(3), right?

Answer 5

yeah right

Answer 6

So, we know that in a first case,
\(\large\color{black}{ f(\color{blue}{2})=\color{blue}{8} }\)

Answer 7

So, now lets go to Part B.

Answer 8

I got disconnected, apologize.

Answer 9

So, lets go to part B.
you want to first find \(\large\color{black}{f^{-1}(x) }\), and then to plug in 3 for x. This will give you the \(\large\color{black}{f^{-1}(3) }\)

In our case, \(\large\color{black}{f(x)=\frac{\LARGE 2x-7}{\LARGE 3} }\)

Answer 10

TO find the inverse function \(\large\color{black}{f^{-1}(x) }\)

you have \(\large\color{blue}{y=\frac{\LARGE 2x-7}{\LARGE 3} }\)

1) switch the \(\large\color{black}{y }\) and the \(\large\color{black}{x }\). ( mean switch them literally)
2) solve for y.

Answer 11

when you are done solving for y (in step 2), substitute/re-write \(\large\color{black}{f^{-1}(x) }\) instead of \(\large\color{black}{y }\). That will be your inverse function, \(\large\color{black}{f^{-1}(x) }\) .

Answer 12

2(3)-7/3
6-7/3
-1/3
what would that be? lol

Answer 13

-3?

Answer 14

wait, not exactly,

\(\LARGE\color{black}{\color{red}{y}=\frac{\LARGE 2\color{royalblue}{x}-7}{\LARGE 3} }\)

\(\LARGE\color{black}{\color{royalblue}{x}=\frac{\LARGE 2\color{red}{y}-7}{\LARGE 3} }\)

\(\large\color{black}{3\color{royalblue}{x}=2\color{red}{y}-7 }\)

\(\large\color{black}{3\color{royalblue}{x}+7=2\color{red}{y} }\)

\(\large\color{black}{\frac{3}{2}\color{royalblue}{x}+\frac{7}{2}=\color{red}{y} }\)

Answer 15

So the inverse function is, \(\large\color{black}{ f^{-1}(x)=\frac{3}{2}x+\frac{7}{2}}\)

Answer 16

oh okay

Answer 17

To find, \(\large\color{black}{ f^{-1}(\color{blue}{3})}\) just plug in \(\large\color{blue}{ 3}\) instead of x.

Answer 18

(into the \(\large\color{black}{ f^{-1}(x)}\) )

Answer 19

f(3) = 3/2(3) + 7/2
=8

Answer 20

Yes.

Answer 21

And the Part C. \(\large\color{black}{ 2y + 14 = 4y - 2}\) can you solve for y?

Answer 22

yes. 2y + 14 = 4y - 2
16y =2y
= 8

Answer 23

yes.

Answer 24

but you meant just 16=2y... not 16 `y`=2y

Answer 25

yeah

Answer 26

thank you for the help

Answer 27

yes, so they are all equivalent:)
You welcome!