## Calculus: Early Transcendentals 8th Edition

(a) $$f\circ g(x)= \ln(x^2-9)$$ Domain: $(-\infty,-3)\cup(3,\infty)$. (b) $$g\circ f(x) =(\ln x)^2-9.$$ Domain: $(0,\infty)$. (c) $$f\circ f(x) =\ln(\ln x).$$ Domain: $(1,\infty)$ (d) $$g\circ g(x)=x^4-18x^2+72.$$ Domain: $(-\infty,\infty)$
(a) $$f\circ g(x)=f(g(x)) = \ln(g(x)) = \ln(x^2-9)$$ The argument of $\ln$ has to be positive so we demand that $x^2-9>0$. This can be rewritten as $(x-3)(x+3)>0$. This is positive either when both $(x-3)$ and $(x+3)$ are positive which is for $x>3$ or when both $(x-3)$ and $(x+3)$ are negative which is for $x0$ so we have for the domain $(0,\infty)$. (c) $$f\circ f(x) =f(f(x))=\ln(f(x))=\ln(\ln x).$$ We demand that the argument of both $\ln$ functions are positive. This means that $x>0$ but also that $\ln x>0$. We know that $\ln x$ is positive when $x>1$ which is a stronger condition from $x>0$ so the domain is $(1,\infty)$ (d) $$g\circ g(x) = g(g(x)) = g(x)^2-9=(x^2-9)^2-9= x^4-18x^2+81-9=x^4-18x^2+72.$$ This is a polynomial function and it is defined for every real number so the domain is $(-\infty,\infty)$