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LenKa [72]
3 years ago
11

Suppose the polynomial f(x) has the following end behavior: as x approaches infinity, f(x) approaches infinity, and as x approac

hes negative infinity, f(x) approaches negative infinity.
Which of the following polynomials could represent f(x)?
There may be more than one correct answer. Select all correct answers.
A. −x3
B. −2x^5−x^3−4x
C. x^3+10x^2−5x+5
D. ​x^2+x−3
E. 7x^5+4x^2
F. x+8

Mathematics
1 answer:
galina1969 [7]3 years ago
6 0

Answer:

C. x³+10x²−5x+5

E. 7x⁵+4x²

F. x+8

Step-by-step explanation:

The graph they described should look something like the one i drew below.

So all you have to do is look at the number with the highest exponent.

If the exponent is an odd number and the x is positive then that function will have an end behavior like the picture i posted.

x³ is good

-x³ is bad because of the negative sign

x² is bad because exponent is even

x⁷ is good

123x¹ is good

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Answer: $8.95 per yard of fabric

Step-by-step explanation:

Define the following:

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7 0
2 years ago
Olga simplifies the expression x + 7x to 7x2. Use the drop-down menus to complete the statements below to explain why Olga's sol
Tcecarenko [31]

Answer: Olga's solution is incorrect

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Step-by-step explanation:


3 0
3 years ago
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Determine whether the statements are True or False. Justify your answer with an explanation.
frez [133]

Problem 1

<h3>Answer: False</h3>

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Explanation:

The notation (f o g)(x) means f( g(x) ). Here g(x) is the inner function.

So,

f(x) = x+1

f( g(x) ) = g(x) + 1 .... replace every x with g(x)

f( g(x) ) = 6x+1 ... plug in g(x) = 6x

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Now let's flip things around

g(x) = 6x

g( f(x) ) = 6*( f(x) ) .... replace every x with f(x)

g( f(x) ) = 6(x+1) .... plug in f(x) = x+1

g( f(x) ) = 6x+6

(g o f)(x) = 6x+6

This shows that (f o g)(x) = (g o f)(x)  is a false equation for the given f(x) and g(x) functions.

===============================================

Problem 2

<h3>Answer: True</h3>

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Explanation:

Let's say that g(x) produced a number that wasn't in the domain of f(x). This would mean that f( g(x) ) would be undefined.

For example, let

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The g(x) function will always produce the output -2 regardless of what the input x is. Feeding that -2 output into f(x) leads to 1/(x+2) = 1/(-2+2) = 1/0 which is undefined.

So it's important that the outputs of g(x) line up with the domain of f(x). Outputs of g(x) must be valid inputs of f(x).

7 0
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sdas [7]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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