Interval notation of function's domain question

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SUMMARY

The domain of the function f(x) = \(\frac{\sqrt{x^2-x-6}}{x-5}\) is determined by analyzing the conditions under which the expression is defined. The critical points are found by solving the inequality \(x^2 - x - 6 \geq 0\), leading to the intervals where the function is valid. The final domain in interval notation is expressed as \((- \infty, -2] \cup [3, 5) \cup (5, +\infty)\) or alternatively as \((- \infty, -2] \cup [3, +\infty) - \{5\}\), explicitly excluding x = 5 due to the denominator.

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  • Familiarity with interval notation
  • Experience with graphing parabolas
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late347
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Homework Statement


find domain
f(x)= ## \frac{\sqrt{x^2-x-6}}{x-5} ##

Homework Equations

The Attempt at a Solution


[/B]
Well, one sees that x ≠5 and then we further investigate the dividend portion where the under-the-root expression ≥ 0

We can graph the parabola, and find the zeroes. Then investigate the intervals and keep in mind also that x ≠ 5 for the original function.

I'm interested in how is this domain described in the interval notation for domain such as with the [ ] brackets

I confirmed by pen-and-paper results with wolfram alpha and the particular type of notation that wolfram alpha uses is as follows.

domain: { x ∈ ℝ : x>5 or 5>x≥3 or x≤-2}
 
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late347 said:

Homework Statement


find domain
f(x)= ## \frac{\sqrt{x^2-x-6}}{x-5} ##

Homework Equations

The Attempt at a Solution


[/B]
Well, one sees that x ≠5 and then we further investigate the dividend portion where the under-the-root expression ≥ 0

We can graph the parabola, and find the zeroes. Then investigate the intervals and keep in mind also that x ≠ 5 for the original function.

I'm interested in how is this domain described in the interval notation for domain such as with the [ ] brackets

I confirmed by pen-and-paper results with wolfram alpha and the particular type of notation that wolfram alpha uses is as follows.

domain: { x ∈ ℝ : x>5 or 5>x≥3 or x≤-2}

Domain: ##(- \infty,-2] \cup [3,5) \cup (5, +\infty)##

Or: ##(- \infty,-2] \cup [3,+\infty) - \{5\}##
 
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