Consider the graphs of f (x) = StartAbsoluteValue x EndAbsoluteValue + 1 and g (x) = StartFraction 1 Over x cubed EndFraction. The composite functions f(g(x)) and g(f(x)) are not commutative, based on which observation?

We have the functions:

f(x) = IxI + 1

g(x) = 1/x^3.

Now, we know that the composite functions do not permute.

How we can prove this?

First, two composite functions are commutative if:

f(g(x)) = g(f(x))

Well, you could use brute force (just replace the values and see if the composite functions are commutative or not)

But i will use a more elegant way.

We can notice two things:

g(x) has a discontinuity at x = 0.

so:

f(g(x)) = I 1/x^3 I + 1

still has a discontinuty at x = 0, but:

g(f(x)) = 1/( IxI + 1)^3

here the denominator is IxI + 1, is never equal to zero.

So now we do not have a discontinuity.

Then the composite functions can not be commutative.

C The domains of f(x) and g(x) are different

Step-by-step explanation:

I got the answer right on Edgenuity

46 dollars is the answer.

step-by-step explanation:

6*6 = 36,         36+10=46

my answer is$ 12

step-by-step explanation:

because 15 -2=12