Question 155654
I'll do the first two to get you started



1)


a) Domain of {{{f(x)=(2)/(x^2+1)}}}


{{{x^2+1=0}}} Set the denominator equal to zero. Remember, you cannot divide by zero.


{{{x^2=-1}}} Subtract 1 from both sides



{{{x=sqrt(-1)}}} or {{{x=-sqrt(-1)}}} Take the square root of both sides



Since the square root of negative 1 is <b>not</b> a real number, this means that no real number will make the denominator equal to zero. So there are no domain restrictions. This means that the domain of {{{f(x)=(2)/(x^2+1)}}} is all real numbers.



So the domain of f(x) in set-builder notation is *[Tex \LARGE \left\{x\|x \in \mathbb{R}\right\}] which is <font size="6">(</font>*[Tex \LARGE -\infty,\infty]<font size="6">)</font> in interval notation



b)


Domain of {{{g(x)=sqrt(x-3)}}}



{{{x-3>=0}}} Set the radicand {{{x-3}}} greater than or equal to zero



{{{x>=3}}} Add 3 to both sides



So any number greater than or equal to 3 is in the domain of {{{g(x)=sqrt(x-3)}}}


So the domain of g(x) in set-builder notation is *[Tex \LARGE \left\{x\|x \ge 3\right\}] which is <font size="6">[</font>*[Tex \LARGE 3,\infty]<font size="6">)</font> in interval notation



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2)


{{{f(x)=(2)/(x^2+1)}}} Start with the first function



{{{f(g(x))=(2)/((sqrt(x-3))^2+1)}}} Plug in {{{g(x)=sqrt(x-3)}}}. In other words, replace each "x" with {{{sqrt(x-3)}}}



{{{f(g(x))=(2)/((x-3)+1)}}} Square {{{sqrt(x-3)}}} to get {{{x-3}}}. Note: the value of {{{x-3}}} is now positive. This means that {{{x>=3}}}



{{{f(g(x))=(2)/(x-2)}}} Combine like terms.



Domain of f(g(x)):


{{{x-2=0}}} Set the denominator equal to zero. Remember, you cannot divide by zero.


{{{x=2}}} Add 2 to both sides


So if {{{x=2}}}, then the whole denominator is zero. So this means that we must exclude the value {{{x=2}}} from the domain. However, we specified earlier that x <b>must</b> be greater than or equal to 3. Since 2 is not greater than or equal to 3, we don't have to worry about restricting this value (as it has been already done)




So the domain of f(g(x)) in set-builder notation is *[Tex \LARGE \left\{x\| x \ge 3 \right\}] which is <font size="6">[</font>*[Tex \LARGE 3,\infty]<font size="6">)</font> in interval notation