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

1 Answer
Alan P.
Oct 26, 2017

Domain: #x > 0#
#(f/g)(x)=sqrt(x^3)" or, equivalently "x^(3/2)#

Explanation:

#(f(x))/(g(x))=(x^2)/sqrt(x)# ...and since
#color(white)("XXXXXX")#division by zero is undefined, and
#color(white)("XXXXXX")#square root of negative numbers are not defined for Real values
#color(white)("XXX")#we can eliminate any values of #x <= 0#
leaving us with a domain of #x > 0#

#x^2/sqrt(x)=x^2xxx^(-1/2)=x^(3/2)# or
#color(white)("XXXXXXXXXX")=(sqrt(x))^3# or
#color(white)("XXXXXXXXXX")=sqrt(x^3)^2#

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