Practice: Functions, Domains and Inequalities

(a) $f(x)={\displaystyle \frac{\ln (x-1)}{x-4}}$. Two operations can fail: - the logarithm requires its argument strictly positive: , so ;
- the division requires a nonzero denominator: $x-4\ne 0$, so $x\ne 4$. Both must hold at once, so the domain is :

The bracket at 1 is open because $\ln 0$ is undefined, unlike a square root, which would have admitted its endpoint. Both brackets at 4 are open because the point is removed entirely, splitting the interval in two. Check: $x=1$ fails the logarithm; $x=1.5$ is admitted; $x=4$ divides by zero; $x=5$ is admitted. (b) $g(x)=\sqrt{9-x^2}$ and $1/g(x)$. For $g$, the root requires $9-x^2\ge 0$, that is $x^2\le 9$:

with closed brackets, since $\sqrt{0}=0$ is perfectly defined and $g(\pm 3)=0$. For $1/g$, the same root condition applies and the denominator must not vanish, so $9-x^2\ne 0$ as well. The two conditions collapse into one strict inequality, :

Why they differ. At $x=\pm 3$ the radicand is exactly 0. The square root accepts that and returns 0, but the reciprocal of 0 does not exist. Placing the root in a denominator therefore converts $\ge$ into and removes precisely the two endpoints. Check: $g(3)=0$, defined; $1/g(3)$ divides by zero. (c) The formula's domain against the model's. The natural domain of $\frac{450}{n}+2.80$ is every real $n\ne 0$: only the division can fail. The model's domain is narrower, because $n$ counts parcels, so it is the positive integers. The model's domain governs whether an output means anything. At $n=-30$ the arithmetic runs without complaint, giving $450/(-30)=-15$ and $-15+2.80=-12.20$, a negative cost per parcel that no spreadsheet will flag. The computation is valid and the answer is meaningless. At $n=30$, by contrast, the model gives $450/30+2.80=17.80$, and the cost falls toward the marginal $2.80$ as $n$ grows. (d) Why the domain is part of the function. Two functions with the same formula and different domains are different functions, because a function is a rule together with the set it applies to. $\sqrt{x}$ on $[0,\mathrm{\infty })$ is surjective onto $[0,\mathrm{\infty })$ and answers "what is the square root of any non-negative number". $\sqrt{x}$ on $[4,9]$ has range $[2,3]$ and answers a different question, one with a bounded answer. The practical consequence is that questions about limits, continuity, differentiability and integrability are questions about points of the domain. Asking whether $f$ is differentiable at a point outside it is not a hard problem but a category error, which is why the domain is settled before any of those questions is posed.

(a) $-3x+1\le 10$. Subtract 1: $-3x\le 9$. Dividing by $-3$ is dividing by a negative, so the direction reverses:

Test a point the answer admits: $x=0$ gives $-3(0)+1=1\le 10$, true. Test one it excludes: $x=-4$ gives $-3(-4)+1=13$, which is not $\le 10$, so the exclusion is right. Had the direction been kept, the answer $x\le -3$ would have excluded 0 and admitted $-4$, and both tests would have failed. Two substitutions settle the question that the rule alone leaves open to a slip. (b) $|3x-9|\ge 6$. An absolute value measures distance from zero, so $|u|\ge 6$ says $u$ is at least 6 away, in either direction. That unfolds to two separate conditions joined by or:

Solving each: $3x\le 3$ gives $x\le 1$; $3x\ge 15$ gives $x\ge 5$. Dividing by 3 is dividing by a positive, so neither direction changes.

The brackets are closed because the inequality is $\ge$: at $x=1$ and $x=5$ the expression equals exactly 6. Why a union. The excluded middle is where the expression is close to zero. At $x=3$ it is $|0|=0$, not $\ge 6$, so the interval between the rays is genuinely absent from the solution set. Writing the answer as $-6\ge 3x-9\ge 6$ would assert that one quantity is simultaneously at most $-6$ and at least 6, which is the empty set. (c) . Here the condition is within 7 of zero, which is a single band and therefore an intersection:

so $(-4,3)$, with open brackets since the inequality is strict. Check: at $x=0$ the expression is , admitted; at $x=3$ it is exactly 7, correctly excluded; at $x=-4$ it is $|-7|=7$, also excluded. Comparing (b) and (c) is the point: the same notation gives a union in one case and an intersection in the other, and which it is depends only on the direction of the inequality. (d) ${\log }_{2}x+{\log }_2(x-2)=3$. Domain first. ${\log }_{2}x$ requires ; ${\log }_2(x-2)$ requires . The stricter governs, so the domain is . Recording this before any algebra is what makes the final check possible, because the combined form will not show these restrictions. Combine and solve. By $\log u+\log v=\log (uv)$, valid here since both arguments are positive on the domain,

which factors as $(x-4)(x+2)=0$, giving candidates $x=4$ and $x=-2$. Check each against the domain. | Candidate | In ? | Verdict |
|---|---|---|
| $x=4$ | yes | solution |
| $x=-2$ | no | discarded, since ${\log }_2(-2)$ does not exist | Verify the survivor: ${\log }_{2}4+{\log }_{2}2=2+1=3$, and $4(4-2)=8=2^3$. (e) Why a valid step produces a false root. The identity $\log u+\log v=\log (uv)$ holds where both sides are defined. The left side requires and separately; the right side requires only that the product be positive, which is also true when both factors are negative. At $x=-2$ the product is , so the rewritten equation is satisfied while the original is meaningless. Nothing went wrong in the algebra. The step enlarged the domain, so the solution set of the rewritten equation can strictly contain that of the original, and the two coincide only after the domain check. That is why stating the domain before solving, and testing every candidate afterwards, belongs to the method rather than being a precaution. Another manipulation with the same property: squaring both sides. From $\sqrt{x+7}=x+1$, squaring gives $x+7=x^2+2x+1$, hence $x^2+x-6=0$ and $x=2$ or $x=-3$. Only $x=2$ works: at $x=-3$ the left side is $\sqrt{4}=2$ while the right is $-2$, and squaring had discarded the sign. Multiplying both sides by an expression that can vanish behaves the same way, introducing roots at its zeros.