The key to understanding how operations with constants effect derivatives
is understanding composition of functions, and operations with constants.
It is often useful to think graphically about operations with constants
(i.e. vertical and horizontal shifts and strecthing of graphs) to get
an idea of what ought to happen to the derivative. For example, a
vertical shift shouldn't change the derivative at all, since the derivative
is measuring slope, which doesn't change when the graph shifts up or down.
By constrast, stretching a graph vertically makes slopes steeper (either
up or down) so the values of the derivative should get bigger.
Such mental pictures should help you remember the content of Theorem 1 in
section 3.2 about how the derivative changes when you do an operation
by constants. For example, in the case of a vertical shift, to make
the values of the derivative bigger, a natural thing to do is to multiply by
the constant doing the stretching (what else is there to multiply by?)
and fortunately, Theorem 1 tells us, this is exactly the right thing to do.
The difficulty with this problem lies in the fact that we have no
formula for F(x) or G(x). However, since we know their derivatives, if
we can recognize pieces of the given derivative which come from their
derivatives via operations by constants, we can back track to find the
desired antiderivative.
- In this case,
is similar to G'. The only
difference is it is shifted 3 units to the right. Thus, if we shift
G(x) three units to the right, its derivative will match up exactly.
Hence, G(x-3) is the desired antiderivative.
- Here,
is made up of one piece like G'
and one piece like F'. First of all, notice that 
. Thus 
is actually F'(x) compressed
horizontally by a factor of 3. However, if we naively try sticking a
3 in the parentheses to horizontally compress F, i.e. F(3x),
Theorem 1 tells us the derivative is 3 F'(3x). That is, the
derivative of F(3x) is compressed horizontally and streteched
vertically. Since we don't want the vertical stretching, a reasonable
thing to do is to cancel it out by first compressing vertically.
Specifically, applying the rules from Theorem 1 shows
is an antiderivative for the first piece.
Similarly, 
is just G'(x) stretched vertically by a
factor of 5. Since the derivative of 5 G(x) is also the derivative
of G(x) stretched veritcally by a factor of 5, it is clear 5 G(x)
is the antiderivative for which we were looking. Combining the two
pieces give the final answer.
- With
, the first piece already matches up with
G'(x). The difficulty lies with the -1, which is there as a trap.
In fact, we know the derivative of x is 1, so altogether the antiderivative
we want is G(x) - x. The point is, shifting the graph of the
derivative down one unit changes the antiderivative. Since we are
used to thinking that shifting a graph vertically doesn't change the
derivative, you may have been lulled into thinking you could shift the
graph of the derivative without effecting the original function. Obviously,
this is not true.