Combining and Composing Functions: Practice

Evaluate each function at $2$, then subtract. Here $f(2) = 5(2) = 10$ and $g(2) = 2 - 1 = 1$.

$(f - g)(2) = 10 - 1 = 9$

The product $fg$ multiplies the two outputs at the same input. Here $f(3) = 3^2 = 9$ and $g(3) = 3$.

$(fg)(3) = 9 \cdot 3 = 27$

Evaluate at $0$ and add. Here $f(0) = 2(0) = 0$ and $g(0) = 0 + 5 = 5$.

$(f + g)(0) = 0 + 5 = 5$

The input being $0$ does not force the output to be $0$.

Work from the inside out. First run $g$ on $3$, then run $f$ on the result. Here $g(3) = 2(3) = 6$.

$f(6) = 6 + 1 = 7$

So $f(g(3)) = 7$. The value $8$ comes from doing the functions in the wrong order.

Composition feeds the output of $g$ into $f$, so the inner function is $g$.

$(f \circ g)(x) = f(g(x))$

The product and sum are different combinations, and $g(f(x))$ is the opposite order.

The inner function here is $f$, so run $f$ on $2$ first. Here $f(2) = 3(2) = 6$.

$g(6) = 6 + 2 = 8$

So $g(f(2)) = 8$. Running them the other way would give $f(g(2)) = f(4) = 12$.

Subtract the two rules at the same input. The $x$ terms cancel.

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

What remains is the constant $7$.

Add the two rules and combine like terms.

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

The constants $-2$ and $+6$ add to $4$.

Evaluate each at $2$, then multiply. Here $f(2) = 2(2) = 4$ and $g(2) = 3(2) = 6$.

$(fg)(2) = 4 \cdot 6 = 24$

Evaluate each at $10$ and subtract. Here $f(10) = 11$ and $g(10) = 9$.

$(f - g)(10) = 11 - 9 = 2$

The ring sign $\circ$ is the composition sign, not multiplication.

$f \circ g \;\text{is read } f \text{ composed with } g$

It means apply $g$ first, then $f$.

First run $g$ on $2$, then run $f$. Here $g(2) = 2 + 4 = 6$.

$f(6) = 10 - 6 = 4$

So $f(g(2)) = 4$.