The function $f$ and $f^2$ are such that $f:x→hx+k$ and $f^2:x→9x+16$

Question

The function $f$ and $f^2$ are such that $f:x→hx+k$ and $f^2:x→9x+16$

please help, im new to functions and dont understand a thing

the answer sheet says $h=3$, $k=4$ ; $h=-3$, $k=-8$

Answer 1

HINT

You define $f(x) = hx+k$ for some fixed real numbers $h$ and $k$ and for all possible $x$. Then, $$ f^2(x) = f(f(x)) = h\left(f(x)\right) + k = h\left(hx+k\right) + k = h^2x + (hk+k) $$

Can you take it from here?

Answer 2

Apparently, here $f^2$ stands for the composition of the function $f$ with itself, i.e. $f^2(x)=(f\circ f)(x)=f(f(x))$. The question tells you that $f(x)=hx+k$ (a linear function) and that $f^2(x)=f(f(x))=9x+16$. But on the other hand, you can evaluate $f^2(x)$ as $$f^2(x)=f(f(x))=f(hx+k)=h(hx+k)+k,$$ which has to be equal to $9x+16$ as a function. So all respective coefficients — the coefficient of $x$ and the constant term — have to be equal in the two expressions for the same function $f^2$. Expanding the expression above, combining the like terms in it, and setting the respective coefficients equal to each other, you'll be able to solve for $h$ and $k$.