How do you find the inverse of #f(x)=4x^2+1, x>=0#? Precalculus Functions Defined and Notation Function Composition 1 Answer kubik98 Jan 3, 2018 #f(x)=4x^2+1;quadquadquadx>=0# #f^(-1)(x)=sqrt(x-1)/2;quadquadquadx>=1# Explanation: #y=4x^2+1# #x=4y^2+1# #x-1=4y^2# #1/4(x-1)=y^2# #+-sqrt(1/4(x-1))=y# #+-sqrt(x-1)/2=y# Answer link Related questions What is function composition? What are some examples of function composition? What are some common mistakes students make with function composition? Is function composition associative? Is it always true that #(f@g)(x) = (g@f)(x)#? If #f(x) = x + 3# and #g(x) = 2x - 7#, what is #(f@g)(x)#? If #f(x) = x^2# and #g(x) = x + 2#, what is #(f@g)(x)#? If #f(x) = x^2# and #g(x) = x + 2#, what is #(g@f)(x)#? What is the domain of #(f@g)(x)#? What is the domain of the composite function #(g@f)(x)#? See all questions in Function Composition Impact of this question 7876 views around the world You can reuse this answer Creative Commons License