009A Sample Final A, Problem 6 - Revision history

MathAdmin at 03:35, 28 March 2015

2015-03-28T03:35:02Z

MathAdmin at 03:34, 28 March 2015

2015-03-28T03:34:34Z

MathAdmin at 03:33, 28 March 2015

2015-03-28T03:33:37Z

MathAdmin at 03:32, 28 March 2015

2015-03-28T03:32:08Z

MathAdmin at 03:28, 28 March 2015

2015-03-28T03:28:54Z

MathAdmin at 03:28, 28 March 2015

2015-03-28T03:28:05Z

MathAdmin at 03:27, 28 March 2015

2015-03-28T03:27:21Z

MathAdmin at 03:26, 28 March 2015

2015-03-28T03:26:28Z

MathAdmin at 05:40, 27 March 2015

2015-03-27T05:40:09Z

MathAdmin at 05:39, 27 March 2015

2015-03-27T05:39:32Z

@@ Line 33: / Line 33: @@
 |<br>&nbsp;&nbsp;&nbsp;&nbsp;  <math>\lim_{x\rightarrow\pm\infty}\frac{\sqrt{4x^{2}+3}}{10x-20}\,\,\cdot\,\,\pm\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}=\lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{\frac{4x^{2}}{x^{2}}+\frac{3}{x^{2}}}}{\frac{10x}{x}-\frac{20}{x}} = \lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{4+\frac{3}{x^{2}}}}{10-\frac{20}{x}}=\pm\frac{2}{10}=\pm\frac{1}{5}</math>
 |-
-|<br>Thus, we have a horizontal asymptote at <math style="vertical-align: -21%;">y=-1/5</math> on the left (as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -2%;">-\infty</math>), and a horizontal asymptote at <math style="vertical-align: -21%;">y=1/5</math> on the right (as <math style="vertical-align: 0%;">x</math> goes to  <math style="vertical-align: -4%;">+\infty</math>).
+|<br>Thus, we have a horizontal asymptote at <math style="vertical-align: -21%;">y=-1/5</math> on the left (as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -2%;">-\infty</math>), and a horizontal asymptote at <math style="vertical-align: -22%;">y=1/5</math> on the right (as <math style="vertical-align: 0%;">x</math> goes to  <math style="vertical-align: -4%;">+\infty</math>).
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 33: / Line 33: @@
 |<br>&nbsp;&nbsp;&nbsp;&nbsp;  <math>\lim_{x\rightarrow\pm\infty}\frac{\sqrt{4x^{2}+3}}{10x-20}\,\,\cdot\,\,\pm\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}=\lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{\frac{4x^{2}}{x^{2}}+\frac{3}{x^{2}}}}{\frac{10x}{x}-\frac{20}{x}} = \lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{4+\frac{3}{x^{2}}}}{10-\frac{20}{x}}=\pm\frac{2}{10}=\pm\frac{1}{5}</math>
 |-
-|<br>Thus, we have a horizontal asymptote at <math style="vertical-align: -23%;">y=-1/5</math> on the left (as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -2%;">-\infty</math>), and a horizontal asymptote at <math style="vertical-align: -23%;">y=1/5</math> on the right (as <math style="vertical-align: 0%;">x</math> goes to  <math style="vertical-align: -3%;">+\infty</math>).
+|<br>Thus, we have a horizontal asymptote at <math style="vertical-align: -21%;">y=-1/5</math> on the left (as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -2%;">-\infty</math>), and a horizontal asymptote at <math style="vertical-align: -21%;">y=1/5</math> on the right (as <math style="vertical-align: 0%;">x</math> goes to  <math style="vertical-align: -4%;">+\infty</math>).
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 29: / Line 29: @@
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}.</math>
 |-
-|<br>This expression is equal to <math style="vertical-align: -3%;">1</math> for positive values of <math style="vertical-align: 0%;">x</math>, and is equal to <math style="vertical-align: -3%;">-1</math> for negative values of <math style="vertical-align: 0%;">x</math>.  Since multiplying <math style="vertical-align: -20%;">f(x)</math> by an expression equal to <math style="vertical-align: -3%;">1</math> doesn't change the limit, we will add a negative sign to our fraction when considering the limit as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -2%;">-\infty</math>.  Thus,
+|<br>This expression is equal to <math style="vertical-align: -2%;">1</math> for positive values of <math style="vertical-align: 0%;">x</math>, and is equal to <math style="vertical-align: -3%;">-1</math> for negative values of <math style="vertical-align: 0%;">x</math>.  Since multiplying <math style="vertical-align: -20%;">f(x)</math> by an expression equal to <math style="vertical-align: -2%;">1</math> doesn't change the limit, we will add a negative sign to our fraction when considering the limit as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -2%;">-\infty</math>.  Thus,
 |-
 |<br>&nbsp;&nbsp;&nbsp;&nbsp;  <math>\lim_{x\rightarrow\pm\infty}\frac{\sqrt{4x^{2}+3}}{10x-20}\,\,\cdot\,\,\pm\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}=\lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{\frac{4x^{2}}{x^{2}}+\frac{3}{x^{2}}}}{\frac{10x}{x}-\frac{20}{x}} = \lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{4+\frac{3}{x^{2}}}}{10-\frac{20}{x}}=\pm\frac{2}{10}=\pm\frac{1}{5}</math>
 |-
-|<br>Thus, we have a horizontal asymptote at <math style="vertical-align: -23%;">y=-1/5</math> on the left (as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -1%;">-\infty</math>), and a horizontal asymptote at <math style="vertical-align: -23%;">y=1/5</math> on the right (as <math style="vertical-align: 0%;">x</math> goes to  <math style="vertical-align: -2%;">+\infty</math>).
+|<br>Thus, we have a horizontal asymptote at <math style="vertical-align: -23%;">y=-1/5</math> on the left (as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -2%;">-\infty</math>), and a horizontal asymptote at <math style="vertical-align: -23%;">y=1/5</math> on the right (as <math style="vertical-align: 0%;">x</math> goes to  <math style="vertical-align: -3%;">+\infty</math>).
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 29: / Line 29: @@
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}.</math>
 |-
-|<br>This expression is equal to <math style="vertical-align: -5%;">1</math> for positive values of <math style="vertical-align: 0%;">x</math>, and is equal to <math style="vertical-align: -5%;">-1</math> for negative values of <math style="vertical-align: 0%;">x</math>.  Since multiplying <math style="vertical-align: -20%;">f(x)</math> by an expression equal to <math style="vertical-align: -5%;">1</math> doesn't change the limit, we will add a negative sign to our fraction when considering the limit as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -2%;">-\infty</math>.  Thus,
+|<br>This expression is equal to <math style="vertical-align: -3%;">1</math> for positive values of <math style="vertical-align: 0%;">x</math>, and is equal to <math style="vertical-align: -3%;">-1</math> for negative values of <math style="vertical-align: 0%;">x</math>.  Since multiplying <math style="vertical-align: -20%;">f(x)</math> by an expression equal to <math style="vertical-align: -3%;">1</math> doesn't change the limit, we will add a negative sign to our fraction when considering the limit as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -2%;">-\infty</math>.  Thus,
 |-
 |<br>&nbsp;&nbsp;&nbsp;&nbsp;  <math>\lim_{x\rightarrow\pm\infty}\frac{\sqrt{4x^{2}+3}}{10x-20}\,\,\cdot\,\,\pm\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}=\lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{\frac{4x^{2}}{x^{2}}+\frac{3}{x^{2}}}}{\frac{10x}{x}-\frac{20}{x}} = \lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{4+\frac{3}{x^{2}}}}{10-\frac{20}{x}}=\pm\frac{2}{10}=\pm\frac{1}{5}</math>
 |-
-|<br>Thus, we have a horizontal asymptote at <math style="vertical-align: -24%;">y=-1/5</math> on the left (as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -1%;">-\infty</math>), and a horizontal asymptote at <math style="vertical-align: -24%;">y=1/5</math> on the right (as <math style="vertical-align: -2%;">x</math> goes to  <math style="vertical-align: 0%;">+\infty</math>).
+|<br>Thus, we have a horizontal asymptote at <math style="vertical-align: -23%;">y=-1/5</math> on the left (as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -1%;">-\infty</math>), and a horizontal asymptote at <math style="vertical-align: -23%;">y=1/5</math> on the right (as <math style="vertical-align: 0%;">x</math> goes to  <math style="vertical-align: -2%;">+\infty</math>).
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 33: / Line 33: @@
 |<br>&nbsp;&nbsp;&nbsp;&nbsp;  <math>\lim_{x\rightarrow\pm\infty}\frac{\sqrt{4x^{2}+3}}{10x-20}\,\,\cdot\,\,\pm\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}=\lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{\frac{4x^{2}}{x^{2}}+\frac{3}{x^{2}}}}{\frac{10x}{x}-\frac{20}{x}} = \lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{4+\frac{3}{x^{2}}}}{10-\frac{20}{x}}=\pm\frac{2}{10}=\pm\frac{1}{5}</math>
 |-
-|<br>Thus, we have a horizontal asymptote at <math style="vertical-align: -25%;">y=-1/5</math> on the left (as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -1%;">-\infty</math>), and a horizontal asymptote at <math style="vertical-align: -25%;">y=1/5</math> on the right (as <math style="vertical-align: 0%;">x</math> goes to  <math style="vertical-align: 0%;">+\infty</math>).
+|<br>Thus, we have a horizontal asymptote at <math style="vertical-align: -24%;">y=-1/5</math> on the left (as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -1%;">-\infty</math>), and a horizontal asymptote at <math style="vertical-align: -24%;">y=1/5</math> on the right (as <math style="vertical-align: -2%;">x</math> goes to  <math style="vertical-align: 0%;">+\infty</math>).
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 7: / Line 7: @@
 |Vertical asymptotes occur whenever the denominator of a rational function goes to zero, <u>''and''</u> &thinsp;it doesn't cancel from the numerator.
 |-
-|On the other hand, horizontal asymptotes represent the limit as ''x'' goes to either positive or negative infinity.
+|On the other hand, horizontal asymptotes represent the limit as <math style="vertical-align: 0%;">x</math> goes to either positive or negative infinity.
 |}
@@ Line 29: / Line 29: @@
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}.</math>
 |-
-|<br>This expression is equal to 1 for positive values of ''x'', and is equal to -1 for negative values of ''x''.  Since multiplying ''f''(''x'') by an expression equal to 1 doesn't change the limit, we will add a negative sign to it when considering the limit as x goes to <math style="vertical-align: -5%;">-\infty</math>.  Thus,
+|<br>This expression is equal to <math style="vertical-align: -5%;">1</math> for positive values of <math style="vertical-align: 0%;">x</math>, and is equal to <math style="vertical-align: -5%;">-1</math> for negative values of <math style="vertical-align: 0%;">x</math>.  Since multiplying <math style="vertical-align: -20%;">f(x)</math> by an expression equal to <math style="vertical-align: -5%;">1</math> doesn't change the limit, we will add a negative sign to it when considering the limit as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -2%;">-\infty</math>.  Thus,
 |-
 |<br>&nbsp;&nbsp;&nbsp;&nbsp;  <math>\lim_{x\rightarrow\pm\infty}\frac{\sqrt{4x^{2}+3}}{10x-20}\,\,\cdot\,\,\pm\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}=\lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{\frac{4x^{2}}{x^{2}}+\frac{3}{x^{2}}}}{\frac{10x}{x}-\frac{20}{x}} = \lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{4+\frac{3}{x^{2}}}}{10-\frac{20}{x}}=\pm\frac{2}{10}=\pm\frac{1}{5}</math>
 |-
-|<br>Thus, we have a horizontal asymptote at ''y'' = -1/5 on the left (as ''x'' goes to <math style="vertical-align: -5%;">-\infty</math>), and a horizontal asymptote at ''y'' = 1/5 as ''x'' goes to  <math style="vertical-align: -5%;">+\infty</math>).
+|<br>Thus, we have a horizontal asymptote at <math style="vertical-align: -23%;">y=-1/5</math> on the left (as <math style="vertical-align: 0%;">x</math> goes to <math style="vertical-align: -1%;">-\infty</math>), and a horizontal asymptote at <math style="vertical-align: -23%;">y=1/5</math> on the right (as <math style="vertical-align: 0%;">x</math> goes to  <math style="vertical-align: -1%;">+\infty</math>).
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 27: / Line 27: @@
 |More work is required here.  Since we need to find the limits at <math style="vertical-align: 0%;">\pm\infty</math>, we can multiply our <math style="vertical-align: -20%;">f(x)</math> by
 |-
-|&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}.</math>
+|<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}.</math>
 |-
-|This expression is equal to 1 for positive values of ''x'', and is equal to -1 for negative values of ''x''.  Since multiplying ''f''(''x'') by an expression equal to 1 doesn't change the limit, we will add a negative sign to it when considering the limit as x goes to <math style="vertical-align: -5%;">-\infty</math>.  Thus,
+|<br>This expression is equal to 1 for positive values of ''x'', and is equal to -1 for negative values of ''x''.  Since multiplying ''f''(''x'') by an expression equal to 1 doesn't change the limit, we will add a negative sign to it when considering the limit as x goes to <math style="vertical-align: -5%;">-\infty</math>.  Thus,
 |-
 |<br>&nbsp;&nbsp;&nbsp;&nbsp;  <math>\lim_{x\rightarrow\pm\infty}\frac{\sqrt{4x^{2}+3}}{10x-20}\,\,\cdot\,\,\pm\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}=\lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{\frac{4x^{2}}{x^{2}}+\frac{3}{x^{2}}}}{\frac{10x}{x}-\frac{20}{x}} = \lim_{x\rightarrow\pm\infty}\pm\frac{\sqrt{4+\frac{3}{x^{2}}}}{10-\frac{20}{x}}=\pm\frac{2}{10}=\pm\frac{1}{5}</math>

@@ Line 5: / Line 5: @@
 ! Foundations: &nbsp;
 |-
-|Vertical asymptotes occur whenever the denominator of a rational function goes to zero, <u>''and''</u> it doesn't cancel from the numerator.
+|Vertical asymptotes occur whenever the denominator of a rational function goes to zero, <u>''and''</u> &thinsp;it doesn't cancel from the numerator.
 |-
 |On the other hand, horizontal asymptotes represent the limit as ''x'' goes to either positive or negative infinity.
@@ Line 19: / Line 19: @@
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>0 = 10x-20 = 10(x-2),</math>
 |-
-|which has a root at ''x'' = 2.  This is our vertical asymptote.
+|which has a root at <math style="vertical-align: 0%;">x = 2.</math>  This is our vertical asymptote.
 |}
@@ Line 25: / Line 25: @@
 !Horizontal Asymptotes: &nbsp;
 |-
-|More work is required here.  Since we need to find the limits at <math style="vertical-align: -5%;">\pm\infty</math>, we can multiply our ''f''(''x'') by
+|More work is required here.  Since we need to find the limits at <math style="vertical-align: 0%;">\pm\infty</math>, we can multiply our <math style="vertical-align: -20%;">f(x)</math> by
 |-
 |&nbsp;&nbsp;&nbsp;&nbsp; <math>\frac{\sqrt{\frac{1}{x^{2}}}}{\,\,\,\frac{1}{x}}.</math>