005 Sample Final A, Question 2
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Question Find the domain of the following function. Your answer should be in interval notation
| Foundations: |
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| 1) What is the domain of ? |
| 2) How can we factor ? |
| Answer: |
| 1) The domain is . The domain of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{x}} is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0, \infty)} , but we have to remove zero from the domain since we cannot divide by 0. |
| 2) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2 - x -2 = (x - 2)(x + 1)} |
| Step 1: |
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| We start by factoring Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2 - x - 2} into Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x - 2)(x + 1)} |
| Step 1: | ||||||||||||
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| Since we cannot divide by zero, and we cannot take the square root of a negative number, we use a sign chart to determine when Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x - 2)(x + 1) > 0} | ||||||||||||
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