Difference between revisions of "005 Sample Final A, Question 4"

Revision as of 18:37, 10 May 2015

Question Find the inverse of the following function $f(x)={\frac {3x}{2x-1}}$

Step 1:
Switch f(x) for y, to get $y={\frac {3x}{2x-1}}$ , then switch y and x to get $x={\frac {3y}{2y-1}}$

Step 2:
Now we have to solve for y: Failed to parse (unknown function "\begin{array}"): {\displaystyle \begin{array}{rcl} x & = & \frac{3y}{2y-1}\\ x(2y - 1) & = & 3y\\ 2xy - x & = & 3y\\ 2xy - 3y & = & x\\ y(2x - 3) & = & x\\ y & = & \frac{x}{2x - 3} }

@@ 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>
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+! Step 2:
 |-
-|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>
 |}