Combining and Composing Functions: Practice

Substitute the whole rule $g(x) = 2x$ into $f$ in place of its input.

$(f \circ g)(x) = f(2x) = 2x + 5$

The outer function only adds $5$ to whatever it receives.

Drop the rule $g(x) = x - 4$ into $f$, then distribute the $3$.

$f(g(x)) = 3(x - 4) = 3x - 12$

The multiplier hits both terms inside the parentheses.

The quotient is undefined where the denominator $g(x)$ equals zero.

$x - 3 = 0 \;\Longrightarrow\; x = 3$

So $3$ must be thrown out of the domain.

The inner function is $f$, so substitute $f(x) = 2x + 1$ into $g$, which subtracts $1$.

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

The $+1$ and the $-1$ cancel.

Substitute the whole rule $g(x) = x + 6$ into $f$, which subtracts $5$, then combine the constants.

$(f \circ g)(x) = (x + 6) - 5 = x + 1$

Substitute $g(x) = x - 3$ into $f$, then expand the square.

$f(g(x)) = (x - 3)^2 = x^2 - 6x + 9$

The square of a binomial keeps the middle term $-6x$.

A fraction has no value when its denominator is zero, and here the denominator is $g(x)$.

$\frac{f(x)}{g(x)} \;\text{is undefined when}\; g(x) = 0$

A zero numerator, by contrast, just makes the quotient equal to $0$.

The outer function cannot take $0$, so exclude the input that makes $g(x) = 0$.

$x + 5 = 0 \;\Longrightarrow\; x = -5$

At $x = -5$ the inner output is $0$, which $f$ rejects.

Subtracting a function from itself leaves nothing.

$(f - g)(x) = (x - 7) - (x - 7) = 0$

Every term cancels.

Now $f$ is inner, so double the output of the square.

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

Compare with $f(g(x)) = 4x^2$: reversing the order changes the result.

First run $g$ on $1$, then square. Here $g(1) = 1 + 4 = 5$.

$f(5) = 5^2 = 25$

So $f(g(1)) = 25$.

Substitute $g(x) = x + 2$ into $f$, distribute, then combine constants.

$f(g(x)) = 3(x + 2) - 2 = 3x + 6 - 2 = 3x + 4$