Question

I need help with a function.

Answer 1

Find each value if: \[f(x) = \frac{ 5 }{x + 2 }\]
5. f(3)

Answer 2

I'm trying to figure out how to plug in the f(3) into the equation.

Answer 3

Think of f(x) as the same as y. f(3) means you need to find y when x is 3.

\[\large f(x) = \frac{ 5 }{x + 2 }\]
Eg. if you were asked to find f(8), you would plug in 8 for x: \[\large f(8) = \frac{ 5 }{8 + 2 } \] \[\large f(8) = \frac{ 1 }{ 2 }\]And that's it, f(8) is 1/2.

ie when x=8, y=1/2

You do NOT divide by 8, you aren't "solving" for f.

Answer 4

\[f(3) = \frac{ 5 }{ 3+2 }\]
\[f(3) = \frac{ 5 }{ 5 }\]
\[f(3) = 1 \]
Did I do this correctlty?

Answer 5

Yes

Answer 6

Your question is ambiguous. State clearly what you want.

Answer 7

Are you sure I did it right?

Answer 8

Yep, that's correct. Functions are easy! :)

Answer 9

I wouldn't have given you a medal AND said yes.

Answer 10

Okay, thanks! :D

Answer 11

You gotta have confidence in your own work, and especially have confidence if someone tells you that you did it right.

Answer 12

What about when it's f(m - 2) instead of f(3) ?

Answer 13

Oh now I understood the question. You've done it correctly.

Answer 14

Plug in (m-2) in place of x.

Answer 15

\[f(m-2) = \frac{ 5 }{ m-2 + 2}\]
Then, it would just be 5 over m...

Answer 16

Yes. But you should always plug it in with parentheses FIRST, like (m-2) not m-2. Then you can drop them after plugging it in

Answer 17

Algebra II is SO confusing lol

Answer 18

The parentheses are especially important if you are plugging into something like
\[f(x) = \frac{1}{x^2 - 1}\] since you want to be sure to square whatever you put in. For example, \[f(m-2) = \frac{1}{(m-2)^2 - 1}\]

Answer 19

\[f(m-2) = \frac{ 5 }{ (m-2) + 2 }\]

Answer 20

But, how do I solve it now?

Answer 21

Now you can drop parentheses. You need them when it's something like jvatsim gave

Answer 22

Okay, so then it looks like this:
\[f(m-2) = \frac{ 5 }{ m }\]

Answer 23

correct!

Answer 24

Is that the answer?

Answer 25

Yes, f(m-2) turns the function into another function.

Answer 26

We plugged in something with a variable ("m") so we should expect to get something with "m"s back.

Answer 27

As one of my teachers told me, "If you plug in f( HAMBURGER ) you would get 5/(HAMBURGER +2)." The point is, you can plug anything into a function.

Answer 28

Levi bc. this unknowed term of x there is in denominator of this fraction allways the first step to make this restrictions of x for what value of x the denominator is zero and in this way this fraction will result undefined - do you know in this case for what value of x will result this fraction undefined ?

Answer 29

@Preetha your opinion please ? ty.

Answer 30

@.Sam.