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Let \displaystyle f(x) = \frac{1}{x-3} and \displaystyle g(x) = \frac{1}{x-7}. Then the domain of f\circ g is equal to all reals except for two values, a and b with a

Guest Sep 15, 2014

Best Answer 

 #1
avatar+92623 
+13

Let \displaystyle f(x) = \frac{1}{x-3} and \displaystyle g(x) = \frac{1}{x-7}. Then the domain of f\circ g is equal to all reals except for two values, a and b with a

 

$$\\Let \displaystyle f(x) = \frac{1}{x-3} \quad and \quad \displaystyle g(x) = \frac{1}{x-7}. \\\\
\mbox{Then the domain of $f\circ g$ is equal to all reals except for two values, a and b with a }$$

 

$$\\f \circ g=\dfrac{1}{\frac{1}{x-7}-3}\\\\
=\dfrac{1}{\frac{1-3(x-7)}{x-7}}\\\\\\
=\dfrac{1}{\frac{-3x+22}{x-7}}\\\\\\
=1 \div \frac{-3x+22}{x-7}\\\\
=1 \times \frac{x-7}{-3x+22}\\\\
=\frac{x-7}{22-3x}\\\\$$

 

NOW     x cannot equal to 7 or 22/3  because you cannot divide by zero!

$$x \in R \quad where \;x\ne7 \;and\;x\ne \frac{22}{3}$$

Melody  Sep 16, 2014
 #1
avatar+92623 
+13
Best Answer

Let \displaystyle f(x) = \frac{1}{x-3} and \displaystyle g(x) = \frac{1}{x-7}. Then the domain of f\circ g is equal to all reals except for two values, a and b with a

 

$$\\Let \displaystyle f(x) = \frac{1}{x-3} \quad and \quad \displaystyle g(x) = \frac{1}{x-7}. \\\\
\mbox{Then the domain of $f\circ g$ is equal to all reals except for two values, a and b with a }$$

 

$$\\f \circ g=\dfrac{1}{\frac{1}{x-7}-3}\\\\
=\dfrac{1}{\frac{1-3(x-7)}{x-7}}\\\\\\
=\dfrac{1}{\frac{-3x+22}{x-7}}\\\\\\
=1 \div \frac{-3x+22}{x-7}\\\\
=1 \times \frac{x-7}{-3x+22}\\\\
=\frac{x-7}{22-3x}\\\\$$

 

NOW     x cannot equal to 7 or 22/3  because you cannot divide by zero!

$$x \in R \quad where \;x\ne7 \;and\;x\ne \frac{22}{3}$$

Melody  Sep 16, 2014

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