Constants for the Domain and Range of a Function We're currently learning about this in high school, and I was looking ahead in the textbook. Since we just had finals and we're on break, I wasn't able to ask anyone about this problem.
It would be great if you could help me with the following problem.
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.
 A: Here are some hints, assuming that the ranges of these functions are totally filled out.


*

*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?

*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!
