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Likurg_2 [28]
3 years ago
14

Determine algebraically whether the function is even, odd, or neither even nor odd.

Mathematics
1 answer:
ololo11 [35]3 years ago
6 0

Answer:

odd

Step-by-step explanation:

Just so you know there are shortcuts for determining if a polynomial function is even or odd. You just to make sure you use that x=x^1 and if you have a constant, write it as constant*x^0 (since x^0=1)

THEN!

If all of your exponents are odd then the function is odd

If all of your exponents are even then the function is even

Now you have -4x^3+4x^1

3 and 1 are odd it is an odd function

This a short cut not the legit algebra way

let me show you that now:

For it to be even you have f(-x)=f(x)

For it be odd you have f(-x)=-f(x)

If you don't have either of those cases you say it is neither

So let's check

plug in -x  -4(-x)^3+4(-x)=-4*-x^3+-4x=-4x^3+-4x

that's not the same so not even

with if we factor out -1 .... well if we do that we get -(4x^3+4x)=-f(x)

so it is odd.

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Answer:  14

<u>Step-by-step explanation:</u>

Consecutive even numbers would be {2, 4, 6, 8, 10, ...}

Let x represent the smallest number, then x + 2 is the middle number and x + 4 is the largest number.

1st number: x

2nd number: x + 2

3rd number: x + 4

1st number + 2nd number + 3rd number = Sum

      (x)        +      (x + 2)       +      (x + 4)      =  48

                                                      3x + 6   =  48

                                                      3x          = 42

                                                        x           = 14

1st number: x           ⇒ x = 14

2nd number: x + 2   ⇒ 14 + 2 = 16

3rd number: x + 4    ⇒ 14 + 4 = 18

<u>Check:</u>

14 + 16 + 18 = 48 \checkmark

 

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