Question on top, answer on bottom. I don't understand the solution though. If x = 4 for the second function, the result is -1...?

2017-08-29 1:58 pm

更新1:

This is a problem from a practice test for calculus 1. I know what domain is, but not sure why -1 is excluded from the domain; I can see that it is impossible for the first function, but it seems that it isn't impossible with the 2nd function, so I would think that the answer should be (-∞, ∞). Am I missing something critical here?

更新2:

My question, simplified: Is the domain really (-∞, -1) ∪ (-1, ∞) or is it just (-∞, ∞)? (And why.)

回答 (3)

[匿名]

2017-08-29 2:04 pm

When x< 0: f(x) = (2x² + 13) / (x² - 1)
When x ≥ 0: f(x) = (5x - 26) / (x + 2)

When x = 4 > 0:
f(4)
= (5*4 - 26) / (4 + 2)
= -6/6
= -1

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Sqdancefan

2017-08-29 2:20 pm

f(-1) is undefined, so -1 is excluded from the domain. There are no other points where the function is undefined.

The domain is (-∞, -1) ∪ (-1, ∞).

參考: https://www.desmos.com/calculator/beh3q2ls61

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cidyah

2017-08-29 9:10 pm

It is given f(x)= (2x^2+13)/(x^2-1) , x < 0
It means x can assume any value below 0
x cannot be -1 as this would result in a zero divide
(2x^2+13)/((x-1)(x+1)) = undefined if x=-1

Therefore, the domain excludes -1
(-∞, -1) ∪ (-1, ∞)

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