MHB What is the Domain of 1/x and 1/[(x - 1)(x + 2)(x - 3)]?

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The domain of the function 1/x is all real numbers except 0, expressed as {x|x cannot be 0}. For the function 1/[(x - 1)(x + 2)(x - 3)], the domain excludes the values where the denominator equals zero, specifically x = 1, -2, and 3. Thus, the correct domain is {x|x cannot be 1, -2, 3}. A typo was noted in the discussion, where -1 was incorrectly mentioned instead of -2. Overall, understanding the domain is crucial for determining where these functions are defined.
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Specify the domain of each variable.

1. 1/x

Here x can be any number EXCEPT for 0.

Domain = {x|x CANNOT be 0}

2. 1/[(x - 1)(x + 2)( x - 3)]

Set each factor to 0 and solve for x individually.

x - 1 = 0

x = 1

x + 2 = 0

x = -2

x - 3 = 0

x = 3

Let D = domain

D = {x|x CANNOT be 1, -1, 3}

Correct?
 
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RTCNTC said:
Specify the domain of each variable.

1. 1/x

Here x can be any number EXCEPT for 0.

Domain = {x|x CANNOT be 0}

2. 1/[(x - 1)(x + 2)( x - 3)]

Set each factor to 0 and solve for x individually.

x - 1 = 0

x = 1

x + 2 = 0

x = -2

x - 3 = 0

x = 3

Let D = domain

D = {x|x CANNOT be 1, -1, 3}

Correct?

right but there is a typo error D = $\{x|x\, cannot\, be \, be\, 1, -2, 3\}$
 
Yes, I made a typo which is a very common mistake using the cell phone keyboard.
 
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