Question
Mathematics
Evaluate $\left(f\circ g\right)\left(x\right)$(f@g)(x) and write the domain in interval notation. Write the answer in the intervals as an integer or simplified fraction.
$f\left(x\right)=\frac{x}{x-1}\phantom{\rule{1em}{0ex}}g\left(x\right)=\frac{13}{{x}^{2}-36}$f(x)=(x)/(x-1)quad g(x)=(13)/(x^(2)-36)
Part: $0/2$0//2
Part 1 of 2
$\left(f\circ g\right)\left(x\right)=$(f@g)(x)=
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To evaluate $\left(f\circ g\right)\left(x\right)$(f@g)(x), we need to first find $g\left(x\right)$g(x) and then evaluate $f$f at $g\left(x\right)$g(x).
Given that $f\left(x\right)=\frac{x}{x-1}$f(x)=(x)/(x-1) and $g\left(x\right)=\frac{13}{{x}^{2}-36}$g(x)=(13)/(x^(2)-36), we have
$\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)=f\left(\frac{13}{{x}^{2}-36}\right)=\frac{\frac{13}{{x}^{2}-36}}{\frac{13}{{x}^{2}-36}-1}.$(f@g)(x)=f(g(x))=f((13)/(x^(2)-36))=((13)/(x^(2)-36))/((13)/(x^(2)-36)-1).
$\begin{array}{rl}\left(f\circ g\right)\left(x\right)& =\frac{\frac{13}{{x}^{2}-36}}{\frac{13}{{x}^{2}-36}-1}\cdot \frac{{x}^{2}-36}{{x}^{2}-36}\\ \\ & =\frac{13}{\left({x}^{2}-36\right)-13}\\ \\ & =\frac{13}{{x}^{2}-49}\\ \\ & =\frac{13}{\left(x-7\right)\left(x+7\right)}\end{array}${:[(f@g)(x)=((13)/(x^(2)-36))/((13)/(x^(2)-36)-1)*(x^(2)-36)/(x^(2)-36)],[],[=(13)/((x^(2)-36)-13)],[],[=(13)/(x^(2)-49)],[],[=(13)/((x-7)(x+7))]:}
Therefore, $\left(f\circ g\right)\left(x\right)=\frac{13}{\left(x-7\right)\left(x+7\right)}$(f@g)(x)=(13)/((x-7)(x+7)).
Now, let’s find the domain of $\left(f\circ g\right)\left(x\right)$(f@g)(x). The function is undefined when the denominator $\left(x-7\right)\left(x+7\right)$(x-7)(x+7) equals zero. This happens when $x=7$x=7 or $x=-7$x=-7. So, the domain of $\left(f\circ g\right)\left(x\right)$(f@g)(x) is all real numbers except for $x=7$x=7 and $x=-7$x=-7.
In interval notation, the domain of $\left(f\circ g\right)\left(x\right)$(f@g)(x) is $\left(-\mathrm{\infty },-7\right)\cup \left(-7,7\right)\cup \left(7,\mathrm{\infty }\right)$(-oo,-7)uu(-7,7)uu(7,oo).
Answer: $\left(f\circ g\right)\left(x\right)=\frac{13}{\left(x-7\right)\left(x+7\right)}$(f@g)(x)=(13)/((x-7)(x+7)) with domain $\left(-\mathrm{\infty },-7\right)\cup \left(-7,7\right)\cup \left(7,\mathrm{\infty }\right)$(-oo,-7)uu(-7,7)uu(7,oo).