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

Latest revision as of 20:12, 21 May 2015

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

Foundations:
1) How do you compose two functions, such as given $f=x^{2}$ and $g=x+1$ , what is f $\circ$ g?
2) When should a point x be in the domain of f $\circ$ g?
Answers:
1) We replace all occurrences of x in f with g, so $f\circ g=(x+1)^{2}$ .
2) A point should be in the domain of f $\circ$ g when it is in the domain of g, and g(x) is in the domain of f.

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
+! Foundations:
 |-
-|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>
+|1) How do you compose two functions, such as given <math>f = x^2</math>&nbsp; and &nbsp; <math>g = x + 1</math>, what is f<math>\circ</math>g?
 |-
-|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.
+|2) When should a point x be in the domain of f<math>\circ</math>g?
 |-
-|c) False. <math>y = x^2</math> does not have an inverse.
+|Answers:
 |-
-|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
+|1) We replace all occurrences of x in f with g, so <math>f\circ g = (x + 1)^2</math>.
 |-
-|e) True.
+|2) A point should be in the domain of f<math>\circ</math>g when it is in the domain of g, and g(x) is in the domain of f.
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+! Step 1:
+|-
+|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:
+|-
+|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>
+|}
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+! Final Answers:
 |-
-|f) False.
+|f <math>\circ</math> g = <math> x </math>, and the domain is <math>[1, \infty)</math>.
 |}

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

Latest revision as of 20:12, 21 May 2015

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