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jenyasd209 [6]
3 years ago
15

Which of the following functions could be f(x) ?

Mathematics
1 answer:
notka56 [123]3 years ago
8 0

Answer:

B.  f(x) = -x^3 - x^2 + 7x - 4

Step-by-step explanation:

For this problem, we want to find the fastest-growing term in our given expressions and equate them when x is - infinite and when x is infinite to see the given trends.

For each of these equations, we will simply take the terms with the highest power and consider those.  The two cases we need to consider is + infinite for x and - infinite for x.  Let's check each of these equations.

Note, any value raised to an even power will be positive.  Any negative value raised to an odd power will be negative.

<u>[A] - x^4</u>

<em>When x is +∞ --> - (∞)^4 -->  f(x) is -∞</em>

<em>When x is -∞ --> - (-∞)^4 --> f(x) is -∞</em>

<em />

<u>[B] - x^3</u>

<em>When x is +∞ --> - (∞)^3 --> f(x) is -∞</em>

<em>When x is -∞ --> - (-∞)^3 --> f(x) is ∞</em>

<em />

<u>[C] 2x^5</u>

<em>When x is +∞ --> 2(∞)^5 --> f(x) is ∞</em>

<em>When x is -∞ --> 2(-∞)^5 --> f(x) is -∞</em>

<em />

<u>[D] x^4</u>

<em>When x is +∞ --> (∞)^4 --> f(x) is ∞</em>

<em>When x is -∞ --> (-∞)^4 --> f(x) is ∞</em>

<em />

Notice how only option B, when looking at asymptotic (fastest-growing) values, satisfies the originally given conditions for the relation of x to f(x).

Cheers.

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Answer:

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Step-by-step explanation:

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c) \lim _{x\to 2}\:\frac{x-2}{x^2-2}=\frac{2-2}{\left(2\right)^2-2}=\frac{0}{4-2}=0

d) \lim _{x\to 11}\:\frac{x^2+6x-187}{x^2+3x-154}=\lim _{x\to 11}\:\frac{\left(x-11\right)\left(x+17\right)}{\left(x-11\right)\left(x+14\right)}=\lim _{x\to 11}\:\frac{\left(x+17\right)}{\left(x+14\right)}=\frac{11+17}{11+14}=\frac{28}{25}

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