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If the composition of two functions is:
1/x-3

What would be the domain restriction? Describe how you found that answer.
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I'm pretty sure the answer would be x≠3, because typically in these scenarios, anything the x on the bottom is being subtracted by has that symbol and the number being subtracted from it. I believe this is because x is = to 3, meaning that - 3 would make it 0 again. Thus, x≠3. But if this isn't true, I'd like someone else to explain it to me, and provide me with the correct answer.

Respuesta :

You are correct! The problems that can arise with a function domain are:

  • Denominators that become zero
  • Even-degree roots with negative input
  • Logarithms with negative or zero input

In this case, you have a denominator, and you don't have roots nor logarithms. This means that your only concern must be the denominator, specifically, it cannot be zero.

And you simply have

[tex]x-3\neq 0 \iff x \neq 3[/tex]

So, the domain of this function includes every number except 3.

Answer:

R-{3}

Step-by-step explanation:

Given is a function

[tex]f(x) = \frac{1}{x-3}[/tex]

The domain is the possible values which x can take

Here on analysing the function we find that f(x) becomes undefined when denominator =0

i.e. [tex]x-3 \neq 0\\x\neq 3[/tex]

There cannot be any other restriction as there is no square root sign.

Hence domain is set of all real numbers except 3

In interval notation domain = [tex](-\infty, 3)U(3,\infty)[/tex]

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