Combining and Composing Functions: Free Response

Ten percent off leaves ninety percent of the price, so $d(p) = 0.9p$. A five percent sales tax multiplies whatever it is charged on by $1.05$, so $t(x) = 1.05x$. The tax is charged on the discounted price, so $t$ is the outer function and $d$ is the inner one.

$(t \circ d)(p) = t(d(p)) = 1.05(0.9p) = 0.945p.$

Substitute $p = 200$ into the combined formula.

$(t \circ d)(200) = 0.945(200) = 189.$

The final price is $189$ dollars.

The combined multiplier from Part A is $0.945$.

$1 - 0.945 = 0.055,$

so the customer pays $94.5\%$ of the original price, a net discount of $5.5\%$. That is not $10\% - 5\% = 5\%$, because the $5\%$ tax rate is never applied to the full original price; it is applied to the smaller, already-discounted amount, so it adds back only $4.5\%$ of the original $p$ instead of $5\%$, leaving a net discount of $5.5\%$ rather than the $5\%$ that subtracting the two rates would give.

For the sum, combine the two rules term by term.

$(f+g)(x) = (4x - 1) + (x + 6) = 5x + 5.$

For the product, distribute every term of one binomial across every term of the other.

$(fg)(x) = (4x - 1)(x + 6) = 4x^2 + 24x - x - 6 = 4x^2 + 23x - 6.$

The quotient is undefined wherever its denominator function equals zero.

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

Since $(f/g)(x) = \dfrac{4x-1}{x+6}$, dividing by zero has no value, so $x = -6$ is thrown out even though $f(-6) = 4(-6) - 1 = -25$ is a perfectly good number.

Look at what each combination asks arithmetic to do.

$(f+g)(x) = f(x) + g(x), \qquad \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}.$

Addition of two real numbers is always defined, no matter what those numbers are, so $(f+g)(x)$ needs no test beyond $x$ already being legal for $f$ and for $g$. Division is not always defined: it fails at a zero denominator. So $(f/g)(x)$ carries an extra test that the sum, difference, and product do not, and any input making $g(x) = 0$ must be excluded even when $f$ is perfectly happy there.

Substitute the whole rule for $g$ into $f$.

$(f\circ g)(x) = f(g(x)) = f(x^2 - 2) = 3(x^2 - 2) + 1 = 3x^2 - 5.$

Substitute the whole rule for $f$ into $g$, and expand the square.

$(g\circ f)(x) = g(f(x)) = g(3x+1) = (3x+1)^2 - 2 = 9x^2 + 6x + 1 - 2 = 9x^2 + 6x - 1.$

The two formulas are not the same polynomial: one has a lead coefficient of $3$ and no $x$ term, the other a lead coefficient of $9$ and a middle term. A quick check at $x=1$ confirms it.

$(f\circ g)(1) = 3(1)^2 - 5 = -2, \qquad (g\circ f)(1) = 9(1)^2 + 6(1) - 1 = 14.$

The reason traces to which function runs last. In $f\circ g$, $f$ is outer, so the final step only scales a square and adds a constant, a mild operation. In $g\circ f$, $g$ is outer, so the final step squares the ENTIRE linear rule $3x+1$, and squaring a sum produces a cross term, $2(3x)(1) = 6x$, that has no counterpart on the other side. That cross term is exactly the extra $6x$ separating the two results.

The three algebraic lines are all sound: substituting $g$ into $f$, squaring the root, and simplifying to $x+3$ are each done correctly. The break is in the conclusion. It reasons from the SIMPLIFIED formula alone, but that formula is only reached after the root inside it is squared away, and squaring a root does not undo the requirement that the root existed in the first place. Before $f$ ever runs, $g(x) = \sqrt{x-1}$ has to produce a real number, which demands

$x - 1 \ge 0 \;\Longrightarrow\; x \ge 1.$

So the domain of $f\circ g$ is $x \ge 1$, not every real number.

Plug $x=0$ into the tidy formula first.

$x + 3 \Big|_{x=0} = 3.$

Now try to compute the same value from the original functions.

$g(0) = \sqrt{0 - 1} = \sqrt{-1},$

which is not a real number, so $f(g(0))$ is never actually reached. The shortcut formula appears to accept $x=0$; the real two-step process does not.

An input to a composition has to survive two separate tests. First, $x$ must be a legal input to the inner function $g$, which needs $x - 1 \ge 0$, that is $x \ge 1$. Second, the OUTPUT $g(x)$ must be a legal input to the outer function $f$; here $f(x) = x^2 + 4$ accepts every real number, so this test excludes nothing further.

$\{x : x \ge 1\} \ \cap\ \{\text{all reals}\} = \{x : x \ge 1\}.$

Both tests together give exactly $x \ge 1$: relying on the simplified formula alone silently dropped the first test.

Substitute the whole inner rule into the outer function in each order, keeping the letters symbolic.

$(f\circ g)(x) = f(g(x)) = a(cx + d) + b = acx + ad + b.$

$(g\circ f)(x) = g(f(x)) = c(ax + b) + d = acx + cb + d.$

Two linear expressions in $x$ are equal for EVERY $x$ exactly when their coefficient of $x$ agrees and their constant term agrees. The coefficient of $x$ is already $ac$ on both sides, so that part is automatic; the whole condition reduces to the constants matching.

$acx + ad + b = acx + cb + d \quad\Longleftrightarrow\quad ad + b = cb + d.$

That single equation is the exact condition, true for every choice of $a,b,c,d$ that satisfies it and false for every choice that does not.

Now test $f(x)=2x+6$, $g(x)=3x+12$, so $a=2,b=6,c=3,d=12$.

$ad+b = 2(12)+6 = 30, \qquad cb+d = 3(6)+12 = 30.$

The two sides agree, so this pair commutes. A direct check confirms it: $(f\circ g)(x) = 2(3x+12)+6 = 6x+30$ and $(g\circ f)(x) = 3(2x+6)+12 = 6x+30$.

Test the new pair against the condition, with $a=c=5$, $b=2$, $d=-3$.

$ad+b = 5(-3)+2 = -13, \qquad cb+d = 5(2)+(-3) = 7.$

The two sides disagree, so this pair does not commute, even though $f$ and $g$ share the slope $5$. A direct check confirms the mismatch: $(f\circ g)(x) = 5(5x-3)+2 = 25x-13$ while $(g\circ f)(x) = 5(5x+2)-3 = 25x+7$.

Sharing a slope only guarantees that the $acx$ terms of the two general formulas agree; it says nothing about whether the constants $ad+b$ and $cb+d$ agree, and those depend on the intercepts $b$ and $d$ as well as the slope. So equal slopes are not sufficient on their own; the full condition is a separate constraint on all four numbers.