Question

i need help with f(x) and g(x) problems /).(\

Answer 1

f(x) = 3x + 5, g(x) = 6x2

Find (fg)(x).

Answer 2

can you please explain how to do this? .-.

Answer 3

dont you mean f(g(x))

Answer 4

no i dont, i know how to find f(g(x)) and g(f(x)) but my book says (fg)(x)

Answer 5

@ganeshie8 @mathmale

Answer 6

hi c: can you help!

Answer 7

(fg)(x) is easy to find actually compared to f(g(x)) and g(f(x))

Answer 8

(fg)(x) = f(x) * g(x)

Answer 9

just multiply both the functions together

Answer 10

f(x) = 3x + 5, g(x) = 6x2

(fg)(x) = f(x) * g(x)
= (3x+5) * (6x^2)
= ?

Answer 11

18x^3+30x^2

Answer 12

\(\large \color{red}{\checkmark}\)

Answer 13

yay c: and then how do you work them backwards
Find f(x) and g(x) so that the function can be described as y = f(g(x)).

y = nine divided by x squared + 2

Answer 14

\(\large y = \frac{9}{x^2}+2\)

like this ?

Answer 15

yeah

Answer 16

or

\(\large y = \left(\frac{9}{x}\right)^2+2\)

Answer 17

which one ?

Answer 18

1st one

Answer 19

oh - i hadn't seen that notation before

Answer 20

yeah and we wont be using it much in higher math...

Answer 21

i guess pre-calc isnt a higher math?

Answer 22

we're given \(\large y = f(g(x)) = \frac{9}{x^2}+2\)

lets cookup the composite functions

Answer 23

pre-calc is a higher math at highschool level :)

Answer 24

try below :

\(\large g(x) = \frac{1}{x^2}\)

\(\large f(x) = 9x+2\)

Answer 25

will they work ?
test them and see...

Answer 26

9(1/x^2) +2

Answer 27

no because it would be 9/9x^2

Answer 28

it wud work right ?

Answer 29

? wouldnt 9 be distributed to both the numerator and denominator?

Answer 30

\(\large g(x) = \frac{1}{x^2}\)

\(\large f(x) = 9x + 2\)


\(\large \implies f(g(x)) = f(\frac{1}{x^2}) = 9(\frac{1}{x^2}) + 2 = \frac{9}{x^2} + 2\)

Answer 31

why doesn't it go to the bottom too though?

Answer 32

\(\large 9(\frac{1}{x^2}) + 2 \)

Answer 33

thats a very good question actually, il take time to explain :)

Answer 34

let me ask u a q

Answer 35

whats the value of below :


\(\large 9(\frac{1}{3^2}) \)

?

Answer 36

9/9=1

Answer 37

Correct,

how did u get 9/9 ?

Answer 38

3^2=9 and 9*1=9

Answer 39

yup ! similarly,


\(\large 9(\frac{1}{x^2}) \)

becomes


\(\large (\frac{9*1}{x^2}) \)

Answer 40

\(\large 9(\frac{1}{x^2}) + 2 \)

becomes



\(\large (\frac{9*1}{x^2}) + 2 \)

Answer 41

which is same as :


\(\large \frac{9}{x^2} + 2 \)

Answer 42

oh okay so x is just a placeholder i always forget that

Answer 43

yup, x is just a placeholder

Answer 44

okay 1 more question
Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.

f(x) = x2 - 3 and g(x) = square root of quantity three plus x
ive got f(g(x))=(sqrt3+x)^2 -3
and since its squared its 3+x-3 =x
but i dont know how to go about g(f(x))=sqrt3+(x^2-3)