Difference between revisions of "005 Sample Final A, Question 3"

Revision as of 09:21, 8 May 2015

Question Find f $\circ$ g and its domain if $f(x)=x^{2}+1\qquad g(x)={\sqrt {x-1}}$

Step 1
First we find the domain of g. Since f $\circ$ g = f(g(x)). So if x is not in the domain of g, it is not in the domain of f $\circ$ g. The domain of g is $[1,\infty )$ .

Step 2
To find f $\circ$ g we replace any occurrence of x in f with g, to yield $({\sqrt {x-1}})^{2}+1=x-1+1=x$

Final Answers
f $\circ$ g = $x$ , and the domain is $[1,\infty )$ .

@@ Line 1: / Line 1: @@
 '''Question ''' Find f <math>\circ</math> g and its domain if <math>f(x) = x^2+1 \qquad g(x)=\sqrt{x-1}</math>
 {| 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>
+|First we find the domain of g. Since f <math>\circ</math> g = f(g(x)). So if x is not in the domain of g, it is not in the domain of f <math>\circ</math> g. The domain of g is <math>[1, \infty)</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.
+|To find f <math>\circ</math> g we replace any occurrence of x in f with g, to yield <math>(\sqrt{x - 1})^2 + 1 = x - 1 + 1 = x </math>
-|-
+|}
-|c) False. <math>y = x^2</math> does not have an inverse.
-|-
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
+! Final Answers
-|-
-|e) True.
 |-
-|f) False.
+|f <math>\circ</math> g = <math> x </math>, and the domain is <math>[1, \infty)</math>.
 |}