Question

Find functions f and g such that the given function h = f ∘ g.

h(x) = (1 + 1/x)^x

I would like a hint.

Answer 1

Nooooooooo

Answer 2

Wait a minute. If it is f(g(x)), then that should mean that every "x" of f(x) is turned into (1 + 1/x).

Answer 3

Why isn't it that the x raised to the power is not turned into (1 + 1/x)?

Answer 4

I was taught that wherever the x is, you have to replace it with the function g(x), if it is f(g(x)). In this case, I am confused because there is an "x" in the power that looks like it didn't get replaced.

Answer 5

what?

Answer 6

hello myininaya!

Answer 7

hey

Answer 8

lots of random answers this evening

Answer 9

earlier too

Answer 10

hey answer this one
i'm having a brain fart

Answer 11

how about
\[g(x)=1+\frac{1}{x}\]
\[h(x)=x^x\]

Answer 12

we could do
\[f(x)=(1+\frac{1}{x})^x, g(x)=x\]
but i'm pretty sure this is not what he wants

Answer 13

i like my answer more

Answer 14

\[h(x)=x^x?\]

Answer 15

do you mean f
that won't work

Answer 16

\[f(g(x))=f(1+\frac{1}{x})=(1+\frac{1}{x})^{1+\frac{1}{x}}\]
not right

Answer 17

\[g(x)=1+\frac{1}{x}\]
\[f(x)=x^x\]
\[f\circ g(x)=f(g(x))=f(1+\frac{1}{x})=(1+\frac{1}{x})^x\]

Answer 18

that is actually what i said at first without thinking

Answer 19

no you didn't replace x with g(x)

Answer 20

i believe i did yes

Answer 21

you didn't replace both

Answer 22

you replaced one of the x's

Answer 23

the base but not the exponent

Answer 24

oooooooooooooooooooooh

Answer 25

yes you are right!

Answer 26

seriously what you have here though is exactly what i said and then he questioned me about it and i was like what and then i realized omg hes right

Answer 27

so try
\[f(x)=x^{\frac{1}{x-1}}\]

Answer 28

bet it works now

Answer 29

i just found the inverse of
\[1+\frac{1}{x}\] and stuck that in the exponent

Answer 30

nice!

Answer 31

merci
but nice of you to notice that i screwed up the first time!

Answer 32

i gotta go eat
:)

gj satellite

Answer 33

I'm confused.

Answer 34

\[h=f(g(x))=f(1+\frac{1}{x})\]
so we need to choose f such that we get
\[f(1+\frac{1}{x})=(1+\frac{1}{x})^x\]
if \[g(x)=1+\frac{1}{x}\] and we are suppose to plug in g(x) wherever we see an x in f(x),
then we need \[h(x)=[g(x)]^{[v(x)]}\]
we need to write v in terms of g
we want \[v(x)=x=> g(x)=1+\frac{1}{v(x)}\]
we need to solve this for v
\[vg=v+1=> vg-v=1=>v(g-1)=1=> v=\frac{1}{g-1}\]
so we have
\[h(x)=f(g(x))=(g(x))^\frac{1}{g(x)-1}\]
so this implies
\[g(x)=1+\frac{1}{x}, f(x)=x^\frac{1}{x-1}\]