Question

please help!! the function f and g are defied by *will post*
which equation is equivalent to **will post**

Answer 1

@jhonyy9 @snowflake0531

Answer 2

Do you get what you are supposed to do?

Answer 3

Like, I know you posted this because you don't know, but like, do you know, like, generally, what the equation is supposed to be, and how it would be simplified

Answer 4

no i dont, functions arent really my thing

Answer 5

Let's take it apart f(2x) multiplied by g(-2x) divided by 2
f(x) = x^2
g(x) = 2x

Answer 6

So, let's first do f(2x)
Literally, this just means, in f(x) = x^2, put x, as 2x
So, what is it?

Answer 7

Substitute x as 2x

Answer 8

... f(2 x 2x)?

Answer 9

._.
f(x) = x^2
Now, it has become
f(2x) = ....?
In, x^2 , substitute x as 2x

Answer 10

2x^2??

Answer 11

Yep, so
it's 2x^2 times g(-2x) over 2
Now, we do g(-2x)
We know that g(x) = 2x
So, in ^^^, plug in x as -2x

Answer 12

2x^2?

Answer 13

g(x) = 2x
g(-2x) =....

Answer 14

x is now -2x
Put that into 2x

Answer 15

-4x^2? i suck at this

Answer 16

First of all, I said wrong earlier, the previous one is 4x^2, not 2x^2

Answer 17

Secondly
relook
x=-2x
g(x) = 2x
Now that x is -2x
You have to substitute it
2(-2x)

Answer 18

-4x^2..

Answer 19

2 times -2x

Answer 20

-4x

Answer 21

Yes
Now multiply
f(2x) with g(-2x)
4x^2 times -4x

Answer 22

-16x^3

Answer 23

Now
what is -16x^3 divided by 2

Answer 24

-8x^3

Answer 25

There's your answer~
Although I have a question lol
Do you understand how I got to that, or were you only answering what I asked you?

Answer 26

no i got it; took my a while to catch on :)

Answer 27

Oh, okay, that's nice then

Answer 28

thank you :))

Answer 29

Yw~!

Answer 30

\(\color{#0cbb34}{\text{Originally Posted by}}\) @artlover03
2x^2??
\(\color{#0cbb34}{\text{End of Quote}}\)
no - not is right

if f(x) = x^2 so f(2x) = (2x)^2 what not equal 2x^2

@snowflake0531 more ATTENTION please !!!

Answer 31

I said it was wrong afterwards

Answer 32

\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531
Yep, so
it's 2x^2 times g(-2x) over 2
Now, we do g(-2x)
We know that g(x) = 2x
So, in ^^^, plug in x as -2x
\(\color{#0cbb34}{\text{End of Quote}}\)
not is right

Answer 33

\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531
First of all, I said wrong earlier, the previous one is 4x^2, not 2x^2
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 34

\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531
Yep, so
it's 2x^2 times g(-2x) over 2
Now, we do g(-2x)
We know that g(x) = 2x
So, in ^^^, plug in x as -2x
\(\color{#0cbb34}{\text{End of Quote}}\)
here you said yep - what mean yes

Answer 35

But afterwards, I said that I was wrong\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531
\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531
First of all, I said wrong earlier, the previous one is 4x^2, not 2x^2
\(\color{#0cbb34}{\text{End of Quote}}\)
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 36

\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531
First of all, I said wrong earlier, the previous one is 4x^2, not 2x^2
\(\color{#0cbb34}{\text{End of Quote}}\)
ohh here

Answer 37

Yea, the above was an oops