Write the given function as the composite of two functions, neither of which is the identity function, f(x) = x http://media.education2020.com/evresources/649-03-02-00-00_files/i0270001.jpg
A composite function is just one function contains another function let s(x) and g(x) be separate functions. Notation of composite functions is, (s o g)(x) is the same as s(g(x)) (g o s)(x) is the same as g(s(x)) for example given s(x) = 1 + x and g(x) = x^2 (s o g)(x) = s(g(x)) = 1 + x^2 and (g o s)(x) = g(s(x)) = (1 + x)^2
you should be able to solve this now
This is actually good practice if you ever get into calculus, as it can make taking derivatives a lot easier.
I'm actually really confused lol. My teacher just threw this at me. But thanks, imma sit here and give some thoughts on it hahaa
It is pretty simple actually dont fret, look at my example, (g o s)(x) = g(s(x)) = (1 + x)^2 see a similarity between this composition and your function?
hint: \[\sqrt{x} = x^{\frac{1}{2}}\]
\[\sqrt[3]{x} = x^{\frac{1}{3}}\] essentially, \[\sqrt[n]{x} = x^{\frac{1}{n}}\] where n is just a number
just to explain this notation just incase you didnt understand it
I will do a few more examples maybe that will help, let, g(x) = 2 and f(x) = x then, (f o g)(x) = f(g(x)) = 2 another example let c(x) = x^(1/2) and d(x) = 1 + x then, (c o d)(x) = c(d(x)) = (1+x)^(1/2)
its like subing a number in but instead you are subing in another function
You just need to think about how this function can be split into two functions
My last example is very very similar to your problem
if you make a guess I can tell you if you are on the right track
Just a final note in this problem you are just doing the reverse you are starting wtih a function then devise what two functions one subbed into the other make it up so, f(x) = (g o s)(x) = g(s(x)) = (x^2 + 2)^(1/3) we need to come up with what g(x) and s(x) are to answer this question
Okay so I don't know if I'm even doing this right but I got something close to f(x) = sqrt(x+2) g(x)= x but something is missing
omg no i forgot the about thr ^(1/3)
yes you did :)
You are close but I feel like a teacher wouldnt give you full marks because you can sub any function into g(x) = x or vice versa and get the same function back. you arent creating two distinct functions from the single function (f o g)(x) = (x + 2)^(1/3)
you are very close though
think of it as an inner and outer function
look at my examples again briefly you are on the right track though
essentially change g(x) = x to the outer function and f(x) to the inner function
for example, in the function (x + 1)^5 x + 1 is the inner function x^5 is the outer function
if you sub x + 1 into x^5 you get the function (x + 1)^5
i'm just confused on how to do the ^(1/3)
its just another exponent nothing special
wait so (sqrt(x+2))^2 ???
if you did that, \[\sqrt{x+2}^{2} = ((x+1)^{\frac{1}{2}})^2 = (x+1)^\frac{2}{2} = x + 1\]
f(x) = (x+2)^(1/3) g(x) = x^2
you need to include the composite how are you subing one function into the other to get the one you are looking for
like (fog)(x) ??
if you did that would you get the function you are looking for??
Another example, again, split into an inner and outer function (5 + x)^(1/5) we see that 5+x is an inner function we see that x^(1/5) is the outer function we sub the inner function into the outer function and we get (5 + x)^(1/5)
this example is pretty much exactly the same as your problem
whaaaaaaaaaaaaa :o so it could also be f(x) = ^(1/3) g(x) = x^2 +2 ????????????
yes!!!
g(x) is your inner function f(x) is your outer function
whaaaaaaaaaaaaaaa :o how am i doing this omg
thanks so much!!
I think you understand?
Yeah, I think I understand lol....
now write the composite
If the outer function is subbed into the inner function you will get the answer wrong, so you have to use composite notation
to show your final answer
(f o g) is right??
yes f(g(x)) is correct, I like to think of it as f is eating g to remember the notation where the circle is a mouth but I'm weird
yasssss ^.^ thank you!!
Join our real-time social learning platform and learn together with your friends!