Question

math problem

Answer 1

Remember

\( (g~of)(x) = g(f(x))\)

Answer 2

so what that means for this question is that you have to replace the entire f(x) equation with the x in the g(x) equation

does that make sense?

Answer 3

yep

Answer 4

because f(x) = 6x - 1

so we have
\( g(f(x)) = g(6x-1)\)

and so plug 6x-1 into the equation of g(x) and simplify it

Answer 5

im not sure

Answer 6

So they told you that
\( g(\color{red}{x}) = 4\color{red}{x}^2 + \color{red}{x}\)

But we need to find out what
\( g(\color{red}{6x-1}) = 4\color{red}{(6x-1)}^2 + \color{red}{(6x-1)}\)

Does that make sense? You just plug it in

Answer 7

but x isnt a number

Answer 8

25x?

Answer 9

25x-5x?

Answer 10

yeah, x isn't a number but our final answer is not a number

it's in terms of x

Answer 11

25x-5x?

Answer 12

az?

Answer 13

20x?

Answer 14

so first

what is (6x-1)^2

remember that
(a-b)^2 = a^2 - 2ab + b^2

Answer 15

36

Answer 16

12

Answer 17

1

Answer 18

36-12+1

Answer 19

you can't lose the x!!

Answer 20

36x^2 - 12x + 1

Answer 21

1296x?

Answer 22

what is that

so now we go from

\( g(\color{red}{6x-1}) = 4\color{red}{(6x-1)}^2 + \color{red}{(6x-1)}\)

\( g(\color{red}{6x-1}) = 4(36x^2 -12x+1) + \color{red}{(6x-1)}\)

Do you see?
We have to now distribute the 4 inside the parenthesis

Answer 23

so what should I do with 4

Answer 24

multiply it inside

\( a(b + c) = ab + ac\)

so multiply it with each term

Answer 25

4 times 6?

Answer 26

each term

4 * 36x^2
4 * 12x
4 * 1

Answer 27

144-48+4?

Answer 28

you have to keep the x^2 and the x

you cannot lose them

Answer 29

144x^2-48x+4

Answer 30

az?

Answer 31

there you go so now we have


\( g(\color{red}{6x-1}) = 144x^2 -48x+4 + 6x-1\)

so last step

what is -48x + 6x

what is 4 - 1

Answer 32

-42x and 3

Answer 33

exactly so your final answer is going to be

\( g(f(x)) = g(6x-1) = \boxed{144x^2 - 42x + 3}\)