Difference between revisions of "005 Sample Final A, Question 2"
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| − | ! Step | + | ! Step 2: |
|- | |- | ||
|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 <math>(x - 2)(x + 1) > 0</math> | |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 <math>(x - 2)(x + 1) > 0</math> | ||
| Line 32: | Line 32: | ||
<td align = "center"><math> x<-1 </math></td> | <td align = "center"><math> x<-1 </math></td> | ||
<td align = "center"><math> x=-1 </math></td> | <td align = "center"><math> x=-1 </math></td> | ||
| − | <td align = "center"><math> -1<x< | + | <td align = "center"><math> -1<x<2 </math></td> |
| − | <td align = "center"><math> x= | + | <td align = "center"><math> x=2 </math></td> |
| − | <td align = "center"><math>x | + | <td align = "center"><math>2<x</math></td> |
</tr> | </tr> | ||
<tr> | <tr> | ||
| − | <td align = "center"><math> | + | <td align = "center"><math> Sign: </math></td> |
<td align = "center"><math> (+) </math></td> | <td align = "center"><math> (+) </math></td> | ||
<td align = "center"><math> 0 </math></td> | <td align = "center"><math> 0 </math></td> | ||
| Line 45: | Line 45: | ||
</tr> | </tr> | ||
</table> | </table> | ||
| + | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | ! Step 3: | ||
| + | |- | ||
| + | | Now we just write, in interval notation, the intervals over which the denominator is positive. | ||
| + | |- | ||
| + | | The domain of the function is: <math>(-\infty, -1) \cup (2, \infty)</math> | ||
| + | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | ! Final Answer: | ||
| + | |- | ||
| + | | The domain of the function is: <math>(-\infty, -1) \cup (2, \infty)</math> | ||
|} | |} | ||
Latest revision as of 20:31, 21 May 2015
Question Find the domain of the following function. Your answer should be in interval notation 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 f(x) = \frac{1}{\sqrt{x^2-x-2}}}
| Foundations: |
|---|
| 1) What 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}{\sqrt{x}}} ? |
| 2) How can we factor 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} ? |
| Answer: |
| 1) The domain 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)} . 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: |
|---|
| 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 2: | ||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 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} | ||||||||||||
|
| Step 3: |
|---|
| Now we just write, in interval notation, the intervals over which the denominator is positive. |
| The domain of the function 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 (-\infty, -1) \cup (2, \infty)} |
| Final Answer: |
|---|
| The domain of the function 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 (-\infty, -1) \cup (2, \infty)} |