MathAdmin: Created page with "'''Question ''' Find f $\circ$ g and its domain if $f(x) = x^2+1 \qquad g(x)=\sqrt{x-1}$ {| class="mw-collapsible mw-collapsed" style = "text-align:lef..."

2015-06-01T01:14:17Z

Created page with "'''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:lef..."

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'''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;"
! Foundations:
|-
|1) How do you compose two functions, such as given <math>f = x^2</math>  and   <math>g = x + 1</math>, what is f<math>\circ</math>g?
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|2) When should a point x be in the domain of f<math>\circ</math>g?
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|Answers:
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|1) We replace all occurrences of x in f with g, so <math>f\circ g = (x + 1)^2</math>.
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|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:
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|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:
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|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:
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|f <math>\circ</math> g = <math> x </math>, and the domain is <math>[1, \infty)</math>.
|}

[[005 Sample Final A|'''<u>Return to Sample Exam</u>''']]

005 Sample Final A, Question 3 - Revision history

MathAdmin: Created page with "'''Question ''' Find f \circ g and its domain if f(x) = x^2+1 \qquad g(x)=\sqrt{x-1} {| class="mw-collapsible mw-collapsed" style = "text-align:lef..."

MathAdmin: Created page with "'''Question ''' Find f $\circ$ g and its domain if $f(x) = x^2+1 \qquad g(x)=\sqrt{x-1}$ {| class="mw-collapsible mw-collapsed" style = "text-align:lef..."