Let #f(x)=x^2# and #g(x)=sqrtx#, how do you find the domain and rules of #(f@g)(x)#?

Question

Sophie Adams · Accepted Answer

Please see below.

The domain of a composition #f @ g# is the part of domain(#g#) for which #g(x)# is in domain(#f#). In this question #g(x) = sqrtx# has domain #[0,oo)#. so the domain of the composition can't be any bigger than that. #f(x) = x^2# has domain all real numbers. So every #g(x)# is in the domain of #f# The domain of #f@g# is #[0,oo)#. The rule #(f@g)(x) = f(g(x))# Replace #g(x)# by #sqrtx# #(f@g)(x) = f(sqrt(x))# For any input #(u)#, #f# gives #(u)^2#, so #f(sqrtx) = (sqrtx)^2# But #(sqrtx)^2 = x# Thus, we conclude with #(f@g)(x) = x# where #x# is in #[0.oo)#

Lincoln Anderson · Answer

undefined To find the domain and rules of $(f@g)(x)$, which represents the composition of functions $f$ and $g$, first, we need to understand that the composition $f@g$ means $f(g(x))$.

Given $f(x) = x^2$ and $g(x) = \sqrt{x}$, we need to find $f(g(x))$.

Substitute $g(x)$ into $f(x)$ to get $f(g(x)) = (g(x))^2$.

Thus, $(f@g)(x) = (\sqrt{x})^2$.

Simplify $(\sqrt{x})^2$ to $x$.

Therefore, $(f@g)(x) = x$.

The domain of $(f@g)(x)$ is the same as the domain of $g(x)$, which is $x \geq 0$ since the square root function is defined only for non-negative real numbers.