# Constants for the Domain and Range of a Function

f(x) is a function with domain [0; 4) and range (-1; 2]. If g(x) = a f(bx + c) + d, for some constants a, b, c, and d, and g(x) has domain (1; 2] and range (3; 7], find the values of a, b, c, and d.

So far, I know that b and c affect the domain of the function, while a and d affect the range.

Once again, I am not looking for a complete answer to this problem, but a few hints or an explanation of the process would definitely help. I appreciate the time you took to read this.

1. No matter what number $y$ we might plug in to $f$, we know that $$-1 < f(y) \leqslant 2.$$ Since the range of $g$ is the set $(3, 7]$, we know that $$3 < a f(y) + c \leqslant 7.$$ Now do a bit of rearranging: $$3 - c < a f(y) \leqslant 7 - c.$$ Now, $a$ cannot be zero because the range of $g$ is not just the set $\{0\}$. Moreover, $a$ cannot be negative because multiplying an inequality $x < y \leqslant z$ by a negative number $n$ reverses the inequalities: $nx > ny \geqslant nz$, or, written the other way, $nz \leqslant ny < nx$, which has the opposite "shape" compared to $x < y \leqslant z$. So $a$ must be positive, and we can divide by it: $$\frac{3 - c}{a} < f(y) \leqslant \frac{7 - c}{a}.$$ See a resemblance?
2. If $x$ is in the domain of $g$, then we know that $$1 < x \leqslant 2.$$ In order to be permitted to plug $bx + c$ into $f$, the number $bx + c$ must be in the domain of $f$, so we must have $$0 \leqslant bx + c < 4.$$ Doing some rearranging similar to the above, $$-c \leqslant bx < 4-c.$$ The "shape" of this inequality is opposite the shape of the inequality $1 < x \leqslant 2$. I think you can go from here!