We will upload all assignment of MCA2 before 28th march
We will upload all assignment of MCA2 before 28th march
MCS013 - Assignment 8(d)
A function is onto if and only if for every y in the codomain, there is an x in the domain such that f(x)=y .
So in the example you give, f:R→R,f(x)=5x+2 , the domain and codomain are the same set: R. Since, for every real number y∈R, there is an x∈R such that f(x)=y , the function is onto. The example you include shows an explicit way to determine which x maps to a particular y , by solving for x in terms of y. That way, we can pick any y , solve for f′(y)=x , and know the value of x which the original function maps to that y .
Side note:
Note that f′(y)=f−1(x) when we swap variables. We are guaranteed that every function f that is onto and one-to-one has an inverse f−1 , a function such that f(f−1(x))=f−1(f(x))=x .
MCS013 - Assignment 8(d)
A function is onto if and only if for every y in the codomain, there is an x in the domain such that f(x)=y .
So in the example you give, f:R→R,f(x)=5x+2 , the domain and codomain are the same set: R. Since, for every real number y∈R, there is an x∈R such that f(x)=y , the function is onto. The example you include shows an explicit way to determine which x maps to a particular y , by solving for x in terms of y. That way, we can pick any y , solve for f′(y)=x , and know the value of x which the original function maps to that y .
Side note:
Note that f′(y)=f−1(x) when we swap variables. We are guaranteed that every function f that is onto and one-to-one has an inverse f−1 , a function such that f(f−1(x))=f−1(f(x))=x .