Composite Functions - Algebra 3

1. Let $f(x)=\frac{x-1}{x+2}$ and $g(x)=2x+1$. Which value is excluded from the domain of $(f\circ g)(x)$?
   Answer: $x=-\frac{3}{2}$ When you see a composition like $(f\circ g)(x)$, the key is to remember that the input to $f$ is not $x$; it is $g(x)$. So you need $g(x)$ to be defined, and then you need $g(x)$ to be in the domain of $f$. Here, $g(x)=2x+1$ is defined for all real $x$, and $f$ is undefined only when its denominator is zero: $$x+2=0 \quad\Rightarrow\quad x=-2.$$ So in the composition, $$f(g(x))=\frac{(2x+1)-1}{(2x+1)+2}=\frac{2x}{2x+3},$$ you must exclude any $x$ where $g(x)=-2$: $$2x+1=-2 \quad\Rightarrow\quad 2x=-3 \quad\Rightarrow\quad x=-\frac{3}{2}.$$ Thus $x=-\frac{3}{2}$ is the excluded value. The choice $x=-2$ is a trap: it is excluded from $f$'s own domain, but that does not matter unless $g(x)$ actually equals $-2$, which happens at $-\frac{3}{2}$, not at $-2$. The choice $x=0$ makes the numerator zero, giving $f(g(0))=f(1)=0$, which is perfectly allowed. The choice $x=\frac{1}{2}$ gives $g\left(\frac12\right)=2$, and $f(2)=\frac14$, so it is also valid. Study tip: for a composition, always substitute $g(x)$ into $f$'s denominator and solve $g(x)=\text{excluded value}$, rather than using $x$ alone.
2. Let $f(x)=\frac{1}{x-1}$ and $g(x)=\frac{x+1}{x}$. Which expression is equal to $(f \circ g)(x)$ and gives its correct domain?
   Answer: $x$, $x\neq 0$ First simplify: $\frac{1}{(x+1)/x-1}=\frac{1}{1/x}=x$. The only restriction is $x\neq 0$ from the definition of $g$; $g(x)=1$ has no solution, so $f$ adds no further restriction. Excluding $x=1$ or $x=-1$ is unnecessary, and $\frac{1}{x}$ is a simplification error.
3. A store applies a 20-dollar-off coupon before adding 7% sales tax. If an item's original price is $p$ dollars, which expression gives the final price $F(p)$?
   Answer: $F(p)=1.07p-21.40$ When you see a coupon applied before tax, think of it as a two-step process: first adjust the price, then apply the multiplier for sales tax. The item starts at $p$ dollars, and the 20-dollar coupon is applied first, so the taxable price is $p-20$. Adding 7% tax means multiplying by $1.07$, giving $F(p)=1.07(p-20)$. Distributing, this becomes $F(p)=1.07p-21.40$, which is the right expression. The choice $F(p)=1.07p-20$ is the most common trap: it applies the coupon after tax, effectively charging tax on the original price and then subtracting 20 from the total. That would be correct only if the store taxed first and then applied the coupon. The choice $F(p)=0.07p-20$ uses only the tax rate instead of the full multiplier $1.07$, so it calculates the tax as a separate amount and then subtracts the coupon, which misses the original price entirely. Finally, $F(p)=p-20$ ignores the sales tax completely. A good strategy for these problems is to identify the order of operations in the wording. If a discount happens before tax, write $(p-\text{discount})$ first, then multiply by $1+\text{tax rate}$. If tax is added first, multiply first and subtract after. Expanding the parentheses shows why the 21.40 appears.
4. Let $f(x)=2x+1$ for $x<0$ and $f(x)=x^2-1$ for $x\ge0$. If $g(x)=2-x$, find $(f\circ g)(2)$.
   Answer: $-1$ Whenever you see a composite function with a piecewise definition, work from the inside out. First evaluate the inner function, then decide which piece of $f$ applies using the value you got, not the original $x$. Here, $g(2)=2-2=0$, so $(f\circ g)(2)=f(0)$. Since $0$ satisfies $x\ge 0$, you must use the second branch: $f(0)=0^2-1=-1$. So the value is $-1$. Now look at the wrong answers to see what they represent. Choosing $1$ means you used the first branch $2x+1$ for $f(0)$, but that branch requires $x<0$, and $0$ is not less than $0$. Choosing $5$ means you evaluated $f(2)$ instead of $f(g(2))$: $2(2)+1=5$, which also uses the wrong branch since $2\ge 0$. Choosing $-3$ comes from reversing the composition or stacking errors: if you incorrectly compute $f(2)=5$, then $g(5)=2-5=-3$, but the composition asks for $f(g(2))$, not $g(f(2))$. A useful habit for piecewise composition: evaluate the inner function first, write down its exact output, and then check which interval that output falls into before choosing a branch. This prevents the most common traps of using the wrong piece or the wrong order.
5. If $(f\circ g)(x)=x^2+4x+7$ and $g(x)=x+2$, which of the following is $f(x)$?
   Answer: $f(x)=x^2+3$ When you see a composite function like $(f\circ g)(x)$, remember it means $f(g(x))$. Here $g(x)=x+2$, so you need a function $f$ such that $f(x+2)=x^2+4x+7$. Let $y=x+2$, so $x=y-2$. Substitute into the composite: $f(y)=(y-2)^2+4(y-2)+7$. Expanding, $y^2-4y+4+4y-8+7=y^2+3$. Since $y$ is just a placeholder, $f(x)=x^2+3$. The choice $x^2+4x+7$ is the original composite itself—it confuses the input $x$ with the input $x+2$. The choice $x^2+7$ misses the constant adjustment from substituting $x=y-2$, likely because it ignores the linear terms. The choice $x^2+4x+3$ also fails: if you try $f(x)=x^2+4x+3$, then $f(x+2)=x^2+8x+15$, not the given composite. Your takeaway: when solving for an outer function, substitute $u=g(x)$, rewrite the composite entirely in terms of $u$, then replace $u$ with $x$. This prevents the common trap of treating the composite expression as already being $f(x)$.
6. Let $f(x)=x^2-1$ and $g(x)=2\sin x$. How many values of $x$ in $[0,2\pi)$ satisfy $(f \circ g)(x)=0$?
   Answer: 4 $(f\circ g)(x)=4\sin^2 x-1$. Setting this to zero gives $\sin x=\pm\frac12$. In $[0,2\pi)$, $\sin x=\frac12$ at $\pi/6$ and $5\pi/6$, and $\sin x=-\frac12$ at $7\pi/6$ and $11\pi/6$, for 4 solutions.
7. Let $f(x)=x^2-3x$ and $g(x)=2x+1$. Find all $x$ for which $(f\circ g)(x)=(g\circ f)(x)$.
   Answer: $x=\frac{-2\pm\sqrt{10}}{2}$ Whenever you see a composition question like this, your first step should be to build each composite separately. Remember $(f\circ g)(x)$ means plug $g(x)$ into $f$, while $(g\circ f)(x)$ means plug $f(x)$ into $g$. Here, $(f\circ g)(x)=(2x+1)^2-3(2x+1)=4x^2-2x-2$, and $(g\circ f)(x)=2(x^2-3x)+1=2x^2-6x+1$. Set them equal: $$4x^2-2x-2=2x^2-6x+1 \Rightarrow 2x^2+4x-3=0.$$ By the quadratic formula, $$x=\frac{-4\pm\sqrt{4^2-4(2)(-3)}}{2(2)} =\frac{-4\pm\sqrt{40}}{4} =\frac{-4\pm 2\sqrt{10}}{4} =\frac{-2\pm\sqrt{10}}{2}.$$ So the correct expression is $\frac{-2\pm\sqrt{10}}{2}$. The choice $-2\pm\sqrt{10}$ is double the correct answer; it treats the fraction $\frac{-4\pm2\sqrt{10}}{4}$ as if only the numerator were divided by $2$. The choice $\frac{-2\pm\sqrt{10}}{4}$ comes from factoring $2$ out of the numerator but then failing to cancel it with the denominator $4$. The choice $\frac{2\pm\sqrt{10}}{2}$ uses $+4$ instead of $-4$ for the $-b$ term in the quadratic formula — a sign error. For your study tip: whenever you get a result involving radicals, always check whether you can simplify the fraction by canceling common factors, and double-check the sign on $b$ in the quadratic formula. Compositions just ask you to substitute carefully — once the two expressions are set equal, it's just a quadratic.
8. Let $f(x)=\frac{1}{x^2}$ and $g(x)=\sqrt{x+2}$. What is the domain of $(f \circ g)(x)$?
   Answer: $(-2,\infty)$ $(f\circ g)(x)=\frac{1}{(\sqrt{x+2})^2}=\frac{1}{x+2}$. The square root requires $x\ge -2$, and the denominator requires $x+2\neq 0$, so $x>-2$. The interval including -2 makes the denominator zero, and the other intervals are not related to the actual restrictions.
9. Let $f(x)=x^2-4$ and $g(x)=\sqrt{x+2}$. How many real solutions does the equation $(f \circ g)(x)=(g \circ f)(x)$ have?
   Answer: 0 Compute $f(g(x))=x-2$ for $x\ge -2$, and $g(f(x))=\sqrt{x^2-2}$ for $x^2\ge 2$. The equation $x-2=\sqrt{x^2-2}$ forces $x\ge 2$. Squaring gives $x^2-4x+4=x^2-2$, hence $x=3/2$, which contradicts $x\ge 2$. So there are no real solutions.
10. Let $f(x)=\sqrt{x}$ and $g(x)=\frac{x^2-4}{x-2}$. What is the domain of $(f \circ g)(x)$?
    Answer: $[-2,2)\cup(2,\infty)$ For $x\neq 2$, $g(x)=x+2$, and at $x=2$ the formula for $g$ is undefined. Since $f$ requires $g(x)\ge 0$, we need $x+2\ge 0$ and $x\neq 2$, giving $[-2,2)\cup(2,\infty)$. Including 2 ignores the hole, excluding -2 is too restrictive, and the reversed interval has the wrong sign.
11. Let $f(x)=\sqrt{9-x^2}$. Which of the following functions $g$ makes $(f \circ g)(x)$ have domain all real numbers?
    Answer: $g(x)=3\cos x$ For $f$ to be defined, $9-(g(x))^2\ge 0$, i.e. $|g(x)|\le 3$. The cosine function multiplied by 3 has range $[-3,3]$, so the composition is defined for every real $x$. The quadratic choices either exceed 3 in absolute value for some $x$ or never enter the interval, so each fails.
12. Let $f(x)=\frac{1}{x+1}$ and $g(x)=\frac{1}{x}$. Which set is the domain of $(f\circ g)(x)$?
    Answer: $x\ne0$ and $x\ne-1$ When you see composition $(f\circ g)(x)$, you're plugging one function into another. The domain is not automatically the domain of the outer function; it's all inputs that are allowed in the inner function and whose outputs are then allowed in the outer function. Start with $g(x)=\frac1x$, which requires $x\ne0$. Now compute $f(g(x))=f\left(\frac1x\right)=\frac{1}{\frac1x+1}$. The outer function $f$ is undefined when its input equals $-1$. So you also need $\frac1x\ne -1$. Solving $\frac1x=-1$ gives $x=-1$. Therefore $x=0$ is excluded by $g$, and $x=-1$ is excluded because it makes $g(x)=-1$,which is outside the domain of $f$. The correct domain is $x\ne0$ and $x\ne-1$. Why the others miss the full picture: the choice $x\ne0$ and$x\ne1$ includes an unnecessary restriction — at $x=1$, $g(1)=1$, then $f(1)=\frac12$, so it's perfectly defined. The choice $x\ne-1$ alone forgets that $g$ itself is undefined at $0$. The choice $x\ne0$ alone misses the composition break at $x=-1$, where $g(-1)=-1$, causing division by zero in $f$. The correct answer must combine both restrictions: the inner function's domain and the outer function's domain applied to the inner output. Strategy: for composition questions, find the inner domain first, then set the inner output equal to whatever makes the outer function undefined, and exclude those $x$-values too. This two-step check prevents missing restrictions.
13. Suppose $f(x)=2x-5$ and $(f \circ g)(x)=6x+1$. What is $(g \circ f)(x)$?
    Answer: $6x-12$ From $2g(x)-5=6x+1$, solve $g(x)=3x+3$. Then $g(f(x))=3(2x-5)+3=6x-12$. The other expressions come from sign errors or composing in the wrong order.
14. Let $f(x)=\frac{x-1}{x+2}$ and $g(x)=\frac{2}{x-3}$. Which expression is equal to $(f \circ g)(x)$, and what is its domain?
    Answer: $\frac{5-x}{2(x-2)}$, domain $x\neq 2,3$ Compute $f(g(x))=\frac{2/(x-3)-1}{2/(x-3)+2}=\frac{5-x}{2(x-2)}$. The domain excludes $x=3$ because $g$ is undefined, and $x=2$ because $g(2)=-2$, where $f$ is undefined. Omitting the $x=3$ restriction, using the opposite numerator sign, or keeping $x-3$ in the denominator are the common errors.
15. Let $f(x)=\sqrt{x+1}$ and $g(x)=x^2-1$. Which statement is true?
    Answer: $(f \circ g)(x)=|x|$ for all real $x$. $(f\circ g)(x)=\sqrt{x^2}=|x|$ for all real $x$. $(g\circ f)(x)=x$ only for $x\ge -1$, not for all real $x$. Thus the true statement is the one saying $f\circ g$ equals $|x|$.
16. Let $f(x)=\frac{1}{x+1}$ and $g(x)=\frac{x-1}{x+2}$. Which set of $x$-values must be excluded from the domain of $(f \circ g)(x)$?
    Answer: $\{-2,-\frac{1}{2}\}$ $g$ is undefined at $x=-2$. Also $f(g(x))$ is undefined when $g(x)=-1$; solving $\frac{x-1}{x+2}=-1$ gives $x=-\frac12$. The excluded set is $\{-2,-\frac12\}$. The set containing $-1$ confuses $f$'s own undefined input with a composition restriction, and $\frac12$ is a sign error.
17. Let $f(x)=3x-1$ and $g(x)=x^2+2$. If $h(x)=(f\circ g)(x)-(g\circ f)(x)$, which expression is equivalent to $h(x)$?
    Answer: $-6x^2+6x+2$ Whenever you see function composition like $(f\circ g)(x)$, read it as $f(g(x))$: apply the inner function first, then plug that entire result into the outer function. On questions like this, the safest process is to compute $f(g(x))$ and $g(f(x))$ separately before subtracting. Here $f(g(x))=3(x^2+2)-1=3x^2+5$. And $g(f(x))=(3x-1)^2+2=9x^2-6x+3$. Subtracting: $(3x^2+5)-(9x^2-6x+3)=-6x^2+6x+2$. So the expression $-6x^2+6x+2$ is correct. The distractor $-6x^2-6x+2$ usually comes from squaring $(3x-1)$ as $9x^2+6x+1$; the cross term inside $g(f(x))$ has the wrong sign, which flips the middle term in the final answer. The choice $6x^2-6x-2$ is exactly what you get if you reverse the order and compute $(g\circ f)-(f\circ g)$ instead. The remaining choice $6x^2+6x+2$ combines that reversal with a sign error in the squared binomial and a constant slip; testing your result at $x=0$ or $x=1$ would reveal it doesn't match. Study tip: When subtracting one composite expression from another, put parentheses around the entire second expression and distribute the minus sign term-by-term. Also remember $(a-b)^2=a^2-2ab+b^2$ — the middle sign is easy to drop.
18. Let $f(x)=\sqrt{x-a}$ and $g(x)=x^2-2$. If the domain of $(f \circ g)(x)$ is $(-\infty,-3]\cup[3,\infty)$, what is the value of $a$?
    Answer: 7 $(f\circ g)(x)=\sqrt{x^2-2-a}$. Its domain is $x^2\ge a+2$. For the boundary to be $\pm 3$, $a+2=9$, so $a=7$. The value 9 is the value of $a+2$, not $a$.
19. Let $f(x)=\frac{1}{x-2}$ and $g(x)=\sqrt{x}$. Which set is the domain of $(f\circ g)(x)$?
    Answer: $x\ge0$ and $x\ne4$ Whenever you see a composition like $(f\circ g)(x)$, remember that the domain must satisfy both functions: the input has to work in $g$, and the output of $g$ has to work in $f$. Start with $g(x)=\sqrt{x}$, so you already need $x\ge 0$. Now compute the composition: $$(f\circ g)(x)=f(\sqrt{x})=\frac{1}{\sqrt{x}-2}.$$ The denominator forces $\sqrt{x}\ne 2$,which means $x\ne 4$. Thus the domain is $x\ge 0$ and $x\ne 4$. The choice that says "$x\ge0$ and $x\ne2$" is a classic trap: it nowhere checks $f$'s restriction on $\sqrt{x}$ instead of on $x$. Actually $x=2$ is fine, since $\sqrt2-2\ne0$. The choice that simply says "$x\ge0$" misses the denominator issue entirely. And the choice that says "$x>0$ and $x\ne4$" incorrectly excludes $x=0$, but $\sqrt0=0$, so $f(0)=1/(0-2)$ is defined. So for composition problems, run the input through the inner function first,then ask whether that result is allowed by the outer function. Check both layers and you will avoid the most common domain mistakes.
20. Let $f(x)=\frac{3}{x-2}$. Which function $g$ satisfies $(f \circ g)(x)=x$ for every $x$ in the domain of $g$?
    Answer: $g(x)=\frac{2x+3}{x}$ Set $\frac{3}{g(x)-2}=x$. Then $g(x)-2=\frac{3}{x}$, so $g(x)=2+\frac{3}{x}=\frac{2x+3}{x}$. The alternatives give $-x$, $\frac{3x}{x+2}$, or a different rational function.

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