Question

Please help, will give medals and fan.
-there are multiple parts to this question. If someone could just show me what to do on one of them that would help so much.

If f(x)=6x-3 and g(x)=11-2x and h(x)=5x^2
a.) (f*g)(x)=?
b.) (f*h)(2)=?
c.) (g*h)(x)=?
d.) (g*f*h)(x)
e.) (g*f)(-2)
f.) (g*g)(x)

Answer 1

is it \[f\times g(x)\] or is it \[f\circ g(x)\] make s a big difference

Answer 2

it has a circle

Answer 3

like the second one you typed

Answer 4

that is what i though

Answer 5

lets do this one \[(f\circ g)(x)\]

Answer 6

the first step is to get it out the the circle notation, and write down what it really means \[f\circ g(x)=f(g(x))\]

Answer 7

that is always the first step, no matter what \(f\) and \(g\) are
the second step is to replace the general \(g(x)\) by the spefic one you have, namely \(11-2x\) so \[f(g(x))=f(11-2x)\]

Answer 8

the third and only abstract step is to replace the \(x\) in \(f(x)\) by \(11-2x\)
think \[f(\spadesuit)=6\spadesuit-3\] so \[f(11-2x)=6(11-2x)-3\]

Answer 9

last step is to clean it up with algebra
distribute
combine like terms
i'll leave that to you

Answer 10

63-12x?

Answer 11

So for #2. You do f(g(2)) which is f(20)? Is that how you start this one? and then 6(20)-3 which is 117

Answer 12

Not sure if I did that right

Answer 13

yes, \(63-12x\) is right

Answer 14

oops it is not \[f(g(2))\] is it? it is \[f(h(2))\]

Answer 15

yes! f(h(2))

Answer 16

your answer is right

Answer 17

Awesome, thank you so much. For c g(h(x)) I did 11-2(5x^2) and got -10x^2+11 and that looks correct to me. I'm a little confused about d though

Answer 18

C is right

Answer 19

D you just got to do it twice \[f(g(h(x))\] is the first step
then replace \(h\) by \(5x^2\)
then ...

Answer 20

oops i had them in the wrong order
it is \[g(f(h(x))\] but that doesn't change the first step \[g(f(5x^2))\]

Answer 21

if you want i will check it

Answer 22

I got 60x^2+17, not sure if I did that right though /:

I did 11-2(6(5x^2)-3)
multiplied 6 by 5x^2 and ended up with 11-2(30x^2-3)
then multiplied 2 by (30x^2-3) and combined like terms to get 60x^2+17

Answer 23

probably right
i would do it one step at a time, may be easier

Answer 24

\[f(h(x))=f(5x^2)=6\times 5x^2-3=30x^2-3\]

Answer 25

then \[g(30x^2-3)=11-2(30x^2-3)\]
think you are off by a minus sign

Answer 26

otherwise good, maybe you just dropped it by mistake \[-60x^2+17\] is what i got

Answer 27

oh I wrote my minus sign down but forgot about sticking with it at the very end!

Answer 28

ok well it looks like you got this pretty quickly

Answer 29

Thank you so much for making this easier (: one more question if you don't mind, for the last one do I just do 11-2(11-2x)?

Answer 30

yes

Answer 31

Okay awesome! I really appreciate it.

Answer 32

my pleasure, glad to help someone who helps themselves