Question

let f(x)=1-2x^2 and g(x) = x+1 compute the following
(g o f)(x) similar to the one I just did but backwords. ???

Answer 1

always start this way
\[g\circ f(x)=g(f(x))\] in other words get rid of the circle

Answer 2

does it change things when they are in a different order?

Answer 3

ok did that

Answer 4

then replace the "inside function" by what you actually have
\[g(1-2x^2)\] in this case

Answer 5

oh yes it matter for sure!

Answer 6

oh

Answer 7

ok got that

Answer 8

now g says to take your input and add 1, so your answer is
\[g(1-2x^2)=1-2x^2+1=2-2x^2\]

Answer 9

think of it this way
\[f(\heartsuit)=1-2\heartsuit^2\]
\[g(\spadesuit)=\spadesuit +1\]

Answer 10

ok

Answer 11

try not to think of the 'x" the variable is unimportant

Answer 12

ok

Answer 13

so if I have (g o f)(-1) I do the same steps as above ?

Answer 14

\[f(\triangle)=1-2\triangle ^2\] now you can fill in the triangle with whatever you want

Answer 15

oh no that means the inverse of the composition, that is harder

Answer 16

oh great

Answer 17

oh wait, is that simply
\[g\circ f(-1)\]?

Answer 18

it is not harder sorry, it is easier

Answer 19

yes

Answer 20

\[g\circ f(-1)\] means
1: find f(-1)
2: find g of the result

Answer 21

ok

Answer 22

\[f(-1)=1-2(-1)^2=1-2=-1\] and
\[g(-1)=-1+1=0\] so
\[g\circ f(-1)=0\]

Answer 23

ok that is what i got

Answer 24

To know why does it have to be (g o f) here's a picture of the thing.

So you'll have to map f first to g, then from g to h. Hence, if we're not using the composite function yet, it'll be this.
f(1)=2
Then g(2)=4
Hence, we can see that we have to map the first function first.
Take note that 2=f(1).
Therefore, g[f(1)]=4
Hence, we write the function as either gf(1)=4 or (g o f)(1)=4.