005 Sample Final A, Question 4 - Revision history

Matt Lee at 03:16, 22 May 2015

2015-05-22T03:16:44Z

Matt Lee at 02:44, 11 May 2015

2015-05-11T02:44:07Z

Matt Lee at 02:43, 11 May 2015

2015-05-11T02:43:45Z

Matt Lee at 02:37, 11 May 2015

2015-05-11T02:37:22Z

Matt Lee: Created page with "'''Question''' Find the inverse of the following function $f(x) = \frac{3x}{2x-1}$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Final Answe..."

2015-05-01T04:08:07Z

Created page with "'''Question''' Find the inverse of the following function <math> f(x) = \frac{3x}{2x-1}</math> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Final Answe..."

New page

'''Question''' Find the inverse of the following function <math> f(x) = \frac{3x}{2x-1}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answers
|-
|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
|-
|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
|-
|c) False. <math>y = x^2</math> does not have an inverse.
|-
|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
|-
|e) True.
|-
|f) False.
|}

@@ Line 24: / Line 24: @@
 ! Final Answer:
 |-
-| y = <math>\frac{x}{2x-3}</math>
+|<math> y = \frac{x}{2x-3}</math>
 |}

@@ Line 11: / Line 11: @@
 |-
 | Now we have to solve for y:
-<math> \begin{array}{rcl}
+::<math> \begin{array}{rcl}
 x & = & \frac{3y}{2y-1}\\
 x(2y - 1) & = & 3y\\
@@ Line 18: / Line 18: @@
 y(2x - 3) & = & x\\
 y & = & \frac{x}{2x - 3}
-</math>
+\end{array}</math>
 |}

@@ Line 2: / Line 2: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answers
+! Step 1:
 |-
-|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
+| Switch f(x) for y, to get <math>y = \frac{3x}{2x-1}</math>, then switch y and x to get <math>x = \frac{3y}{2y-1}</math>
 |-
-|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
+| Now we have to solve for y:
-|-
+<math> \begin{array}{rcl}
-|c) False. <math>y = x^2</math> does not have an inverse.
+x & = & \frac{3y}{2y-1}\\
-|-
+x(2y - 1) & = & 3y\\
-|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
+xy - x & = & 3y\\
-|-
+xy - 3y & = & x\\
-|e) True.
+y(2x - 3) & = & x\\
-|-
+y & = & \frac{x}{2x - 3}
-|f) False.
+</math>
 |}