Given $f(x)= x-3$, $ h(x) = x^3$ , $j(x) = 2x$ . Express $y=2x-3$, $y=x^{3/2}$ as a composite involving $f, g, h, j$
How do I combine for example $y=2x-3$ into $f(x)=x-3$
Is it $f(x)= (2x-3)-3$ ? how do I present it as a composite function?
Given $f(x)= x-3$, $ h(x) = x^3$ , $j(x) = 2x$ . Express $y=2x-3$, $y=x^{3/2}$ as a composite involving $f, g, h, j$
How do I combine for example $y=2x-3$ into $f(x)=x-3$
Is it $f(x)= (2x-3)-3$ ? how do I present it as a composite function?
The composition $f\circ g$ of two functions $f$ and $g$ uses the output of $g$ as input for $f$. That is, if we evaluate $f\circ g$ in the point $x$, we first feed $x$ into $g$. This gives us a result $g(x)$, which we then feed into $f$. The final output is $f(g(x))$.
In this case, we want to obtain (to take your first example) $2x-3$ as the final output for the input $x$. That is, we want to double $x$ and subtract $3$ from the result. The function $j$ doubles its input and the function $f$ subtracts $3$ from its input. Therefore, if we first apply $j$ and then $f$ to $x$, we will have doubled $x$ and subtracted $3$. Indeed, $f\circ j(x)=f(j(x))=f(2x)=2x-3$, as desired.
Note the perhaps slightly confusing order. We write $f\circ j$ for the function which first applies $j$ and then applies $f$ to the input.
Hopefully, you can now make $y=x^{3/2}$ yourself.