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 y $y$ in the codomain, there is an x $x$ in the domain such that f(x)=y $f(x)=y$.

**if and only if for every***onto*So in the example you give, f:R→R,f(x)=5x+2 $f:\mathbb{R}\to \mathbb{R},\phantom{\rule{1em}{0ex}}f(x)=5x+2$, the domain and codomain are the same set: R. $\mathbb{R}.\phantom{\rule{thickmathspace}{0ex}}$ Since, for every real number y∈R, $y\in \mathbb{R},\phantom{\rule{thinmathspace}{0ex}}$ there is an x∈R $\phantom{\rule{thinmathspace}{0ex}}x\in \mathbb{R}\phantom{\rule{thinmathspace}{0ex}}$ such that f(x)=y $f(x)=y$, the function is onto. The example you include shows an explicit way to determine which x $x$ maps to a particular y $y$, by solving for x $x$ in terms of y. $y.$ That way, we can pick any y $y$, solve for f′(y)=x ${f}^{\prime}(y)=x$, and know the value of x $x$ which the original function maps to that y $y$.

Side note:

Note that f′(y)=f−1(x) ${f}^{\prime}(y)={f}^{-1}(x)$ when we swap variables. We are guaranteed that every function f $f$ that is onto and one-to-one has an inverse f−1 ${f}^{-1}$, a function such that f(f−1(x))=f−1(f(x))=x $f({f}^{-1}(x))={f}^{-1}(f(x))=x$.

## MCS013 - Assignment 8(d)

A function is y $y$ in the codomain, there is an x $x$ in the domain such that f(x)=y $f(x)=y$.

**if and only if for every***onto*
So in the example you give, f:R→R,f(x)=5x+2 $f:\mathbb{R}\to \mathbb{R},\phantom{\rule{1em}{0ex}}f(x)=5x+2$, the domain and codomain are the same set: R. $\mathbb{R}.\phantom{\rule{thickmathspace}{0ex}}$ Since, for every real number y∈R, $y\in \mathbb{R},\phantom{\rule{thinmathspace}{0ex}}$ there is an x∈R $\phantom{\rule{thinmathspace}{0ex}}x\in \mathbb{R}\phantom{\rule{thinmathspace}{0ex}}$ such that f(x)=y $f(x)=y$, the function is onto. The example you include shows an explicit way to determine which x $x$ maps to a particular y $y$, by solving for x $x$ in terms of y. $y.$ That way, we can pick any y $y$, solve for f′(y)=x ${f}^{\prime}(y)=x$, and know the value of x $x$ which the original function maps to that y $y$.

Side note:

Note that f′(y)=f−1(x) ${f}^{\prime}(y)={f}^{-1}(x)$ when we swap variables. We are guaranteed that every function f $f$ that is onto and one-to-one has an inverse f−1 ${f}^{-1}$, a function such that f(f−1(x))=f−1(f(x))=x $f({f}^{-1}(x))={f}^{-1}(f(x))=x$.