\((H \circ M)(k) = H(M(k)) = H(3 k - 5)\)

\(H\) is going to do something to \(3 k - 5\) and produce \(5 k + 1\).

These are both linear functions, therefore \(H(d)\) should be a linear function.

First \(H\) should multiply \(3 k - 5\) by \(\answer {\frac {5}{3}}\) to produce a \(5k\).

Now, we need \(-\frac {25}{3}\) to become \(1\). \(H\) should add \(\answer {\frac {28}{3}}\).

\(\blacktriangleright \) Check:

\((H \circ M)(k) = H(M(k)) = 3 M(k) - 5 \)

These are both linear functions, therefore \(M(k)\) should be a linear function.

\(M\) is going to be multiplied by \(3\), therefore \(M\) should start with \(M(p) = \answer {\frac {5}{3}}p\). That way, when it is multiplied by \(3\), we will get \(5p\).

Now, we need \(3\) times something to be \(1\). \(M\) should add \(\answer {\frac {1}{3}}\).

\(\blacktriangleright \) Check:

\((f \circ g)(t) = f(g(t)) = \frac {2t}{t-1}\)

The composition has two occurrences of the variable. \(g\) has one occurrence and when it is put into \(f\) the resulting formula has two occurrences. It seems reasonable to guess that \(f\) has two occurences of its variable - one in a numerator and one in a denominator. \(3t + 1\) will replace each in the composition.

So, let’s start constructing.

\(\blacktriangleright \) Guess #1)

Let’s start with two occurences of \(x\), one in the numerator and one in the denominator.

this makes the composition look like

We need the coefficients of \(t\) to be \(2\) in the numerator and \(1\) in the denominator.

\(\blacktriangleright \) Guess #2)

That gives

Now, we need constant terms of \(0\) in the numerator and \(-1\) in the denominator. Subtract \(\answer {\frac {2}{3}}\) from the numerator. Subtract \(\answer {\frac {4}{3}}\) from the denominator.

\(\blacktriangleright \) Guess #3)

That tells us what \(f\) needs to do.

\((B \circ H)(t) = B(H(t)) = 3 H(t) + 1\)

We want this to equal \(\frac {2t}{t-1}\).

We can solve for \(H(t)\).

Check: