Let #f(x)=8x # and #g(x)=x/8#, how do you find each of the compositions and domain and range?

Question

Cora Alexander · Accepted Answer

undefined To find the composition of $f \circ g$ (denoted as $f(g(x))$), substitute the expression for $g(x)$ into $f(x)$. Similarly, to find the composition of $g \circ f$ (denoted as $g(f(x))$), substitute the expression for $f(x)$ into $g(x)$.

$f(g(x)) = f\left(\frac{x}{8}\right) = 8\left(\frac{x}{8}\right) = x$

$g(f(x)) = g(8x) = \frac{8x}{8} = x$

For the domain and range:

For $f(x) = 8x$, the domain is all real numbers ($\mathbb{R}$) because there are no restrictions on the values $x$ can take. The range is also all real numbers ($\mathbb{R}$) because for any input $x$, $f(x)$ produces an output that spans the entire real number line.

For $g(x) = \frac{x}{8}$, the domain is again all real numbers ($\mathbb{R}$), and the range is also all real numbers ($\mathbb{R}$) because dividing any real number by 8 still results in a real number.

James Allen · Answer

#f(g(x))=xcolor(white)("XXX")#and#color(white)("XXX")g(f(x))=x#
Both functions and their compositions have Domains and Ranges of #(-oo,+oo)# Sometimes the use of #x# in multiple definitions can cause confusion, so let's re-write the base equations as: #color(white)("XXX")f(a)=8a# and #color(white)("XXX")g(b)=b/8# So, replacing #a# with #g(b)# we have #color(white)("XXX")f(g(b)) = 8*g(b) = 8*b/8 = b# (or, using the original #x# as the variable: #f(g(x))=x#) Similarly #g(f(x)) = x# #{ (f(x) = 8x), (g(x) = (x)/(8)), (f(g(x))=x), (g(f(x)))=x :}# #color(white)("XXX")#are all defined for all values of# x# and therefore have Domains of all Real values, #(-oo,+oo)# Replacing, for example #f(x)# with #y# we can see that #f(x)=8x <=>y/8=x# which is defined for all Real values of #y#. (this process can be done for all of the functions) So the Ranges of these functions is also all Real values #(-oo,+oo)#