Poins on Inverse functions

Question

I have tried to do this but, I don't know what to do. They have given me a function y=h(x) that passes through point (3,2) and (4,5). If the function h has a inverse called j, what points must be on the graph of y=j(x). Also, this not homework, I am just preparing for next year. Can someone help me solve it. Show you work otherwise I can't understand by just looking at an awnsers.

Answer 1

Edit: It looks like you know that $(2,3)$ and $(5,4)$ will be points on the graph $y=j(x),$ and just aren't sure why. Let me try to explain it for you.

One way to think of functions is as predictable processes, such that if we give a function $h$ a certain input $x$, we'll always get a particular output, which we call $h(x)$.

Functions are said to be inverses if they "undo" each other's processes. Suppose $h$ and $j$ are inverse functions. Then given an $x$ in the domain of $h$ (meaning $x$ is a valid input for $h$), we know that $h$ will take the input $x$ and gives the output $h(x)$. Well, $j$ undoes $h$'s process, so it takes $h(x)$ as an input, and gives the output $x$--in function notation, $j\bigl(h(x)\bigr)=x.$ This is true for all $x$ in the domain of $h$. Likewise, for all $x$ in the domain of $j,$ we have $h\bigl(j(x)\bigr)=x$.

Now, let's look at this particular example. We can only be sure that: $h$ takes the input $3$ and gives the output $2$; $h$ takes the input $4$ and gives the output $5$. This may well be all that $h$ does--perhaps no other inputs are valid, or perhaps there are many, but we can't say from the information given. However, we can still draw the conclusion (from the fact that $j$ is the inverse of $h$) that: $j$ takes the input $2$ and gives the output $3$; $j$ takes the input $5$ and gives the output $4$. In other words, $(3,2)$ and $(5,4)$ are points on the graph $y=j(x)$. Again, we can't conclude anything else. This is the best we can do, since we only know two points on the graph $y=h(x)$.

More generally, if $(s,t)$ is a point on the graph $y=h(x)$ and $j$ is the inverse function of $h,$ then $(t,s)$ is a point on the graph $y=j(x)$.