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makvit
2 months ago
5

Consider the graphs of f (x) = StartAbsoluteValue x EndAbsoluteValue + 1 and g (x) = StartFraction 1 Over x cubed EndFraction. T

he composite functions f(g(x)) and g(f(x)) are not commutative, based on which observation?
Mathematics
2 answers:
AnnZ [12.3K]2 months ago
7 0

Answer:

We have defined functions:

f(x) = IxI + 1

g(x) = 1/x^3.

Currently, it is evident that the composite functions are not commutative.

How can we demonstrate this?

To determine if two composite functions are commutative, the following must hold true:

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

One could apply brute force (simply substituting values to see if the composite functions commute),

but I will opt for a more sophisticated approach.

There are two notable observations:

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

Thus:

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

remains discontinuous at x = 0, whereas:

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

shows that the denominator IxI + 1 can never reach zero.

At this point, there is no discontinuity.

Consequently, the composite functions cannot be commutative.

Leona [12.6K]2 months ago
4 0

Answer: C The domains of f(x) and g(x) differ

Step-by-step explanation:

I got the answer right on Edgenuity

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Respuesta:

1/16   2/16  3/16  4/16  5/16  6/16  7/16  8/16 9/16 10/16  11/16 12/16

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Para identificar las brocas existen dos aspectos a considerar:

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9/16   3/16    7/16     5/16   11/16    el orden sería (de menor a mayor)

3/16     5/16     7/16     9/16     11/16

2.- Fracciones con diferentes denominadores pueden convertirse a un denominador común /16 multiplicando la fracción, por ejemplo

1/4   =  1*4/4*4   =  4/16

Aplicando este método, todas las fracciones se transforman al formato mencionado previamente y se organizan

1/4   = 4/16

3/8  = 6/16

1/2  = 8/16

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1/8  =  2/16

Por lo tanto, hay diez brocas, comenzando con 1/16 hasta la número 12 que es 12/16

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1/16   2/16  3/16  4/16  5/16  6/16  7/16  8/16 9/16 10/16  11/16 12/16

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1) Initially, there are 9,000 bees in the first year.

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3) Each subsequent year sees a 5% decline => 9,000 * (0.95)^(number of years)

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<span>1) The function f(x) = 9,000(1.05)x applies to the scenario.

FALSE: WE ESTABLISHED IT AS f(x) = 9,000 (0.95)^x

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TRUE: FRACTIONS OF BEES CANNOT EXIST.
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