Combining and Composing Functions: Practice

Compose $f$ with itself: substitute the rule $x + 1$ into $f$, which adds another $1$.

$f(f(x)) = (x + 1) + 1 = x + 2$

Here $(f \circ g)(x) = \dfrac{1}{x - 6}$, and $f$ cannot take $0$, so exclude where $g(x) = 0$.

$x - 6 = 0 \;\Longrightarrow\; x = 6$

At $x = 6$ the inner output is $0$, which $f$ rejects.

Now $f$ runs first, and $f(x) = \dfrac{1}{x}$ rejects $0$, while $g$ accepts everything.

$(g \circ f)(x) = \frac{1}{x} - 6, \quad x \neq 0$

So the domain is all numbers except $0$. Notice the excluded input changed from $6$ to $0$ when the order flipped.

Run $g$ on $3$ first, then apply $f$. Here $g(3) = 3 - 1 = 2$.

$f(2) = 2^2 + 1 = 5$

So $f(g(3)) = 5$.

Substitute $g(x) = x + 1$ into $f$ and expand the square.

$f(g(x)) = (x + 1)^2 = x^2 + 2x + 1$

First build the composition, then find where its denominator is zero.

$f(g(x)) = \frac{1}{(x + 1) - 2} = \frac{1}{x - 1}$

The denominator is zero at $x = 1$, so that input is excluded.

The inner function is $f$, so substitute $2x - 1$ into $g$ and distribute.

$g(f(x)) = 3(2x - 1) = 6x - 3$

The inner function is $h$, so subtract $5$ from the output of the square.

$k(h(x)) = x^2 - 5$

Running the other order would give $h(k(x)) = (x - 5)^2$, a different function.

Build the composition first, then set it equal to $10$. Here $f(g(x)) = (2x - 1) + 3 = 2x + 2$.

$2x + 2 = 10 \;\Longrightarrow\; 2x = 8 \;\Longrightarrow\; x = 4$

Each application of $f$ adds $2$, and there are three of them.

$f(f(f(x))) = x + 2 + 2 + 2 = x + 6$

Compose $f$ with itself by substituting the whole rule $f(x) = \dfrac{1}{x + 1}$ into $f$ wherever it expects its input.

$f(f(x)) = \frac{1}{f(x) + 1} = \frac{1}{\dfrac{1}{x + 1} + 1}$

Write the denominator over a common denominator, $\dfrac{1}{x + 1} + 1 = \dfrac{x + 2}{x + 1}$, then take the reciprocal of that fraction.

$f(f(x)) = \frac{x + 1}{x + 2}$

Substitute $g(x) = x + 1$ into $f$, expand the square, then simplify.

$f(g(x)) = (x + 1)^2 - 1 = x^2 + 2x + 1 - 1 = x^2 + 2x$