Functions, Domains and Inequalities

A polynomial, $f(x)=x^2-4x+7$. Domain $(-\mathrm{\infty },\mathrm{\infty })$. Nothing can fail. Its range is another matter: completing the square gives $(x-2{)}^2+3$, so the minimum is 3 at $x=2$ and the range is $[3,\mathrm{\infty })$. Checking $f$ at $0,1,2,3,4$ gives $7,4,3,4,7$, and a scan of 200,000 random inputs found nothing below.

Domain and range are different questions: the first asks what may go in, the second what comes out.

A rational function, ${\displaystyle \frac{1}{x-3}}$. One division, so one condition: $x\ne 3$. The domain is $(-\mathrm{\infty },3)\cup (3,\mathrm{\infty })$. At $x=2.5$ the value is $-2$; at $x=3$ the computation raises an error rather than producing a large number.

A square root, $\sqrt{x-2}$. Domain $[2,\mathrm{\infty })$, with the bracket at 2 because $\sqrt{0}=0$ is perfectly defined. The restriction is $\ge$, not . A distinction the bracket records.

A logarithm, $\ln x$. Domain $(0,\mathrm{\infty })$, parenthesis at 0 because $\ln 0$ is undefined. Here the restriction is strict where the square root's was not, which is why the two cannot be treated as the same kind of condition.

Both at once, ${\displaystyle \frac{\sqrt{x-2}}{x-5}}$. The root gives $x\ge 2$, the denominator gives $x\ne 5$, and the domain is $[2,5)\cup (5,\mathrm{\infty })$. Testing: $1.9$ is out, $2$ is in, $4.9$ is in, $5$ is out, $5.1$ is in.

A root in a denominator, ${\displaystyle \frac{1}{\sqrt{4-x^2}}}$. Two conditions collapse into one strict inequality: the root needs $4-x^2\ge 0$ and the division needs it $\ne 0$, so , giving $(-2,2)$. Both endpoints are excluded, and computing at $x=2.5$, $3$ or $5$ raises an error.

A composition, $\sqrt{x-1}$ from $f(x)=x-1$ and $g(u)=\sqrt{u}$. $f$ is total and $g$ needs a non-negative input, so the composite needs $x\ge 1$. At $x=0.5$ the computation fails; at $x=1$ it gives 0. The restriction belongs to neither original domain.

Three operations, each contributing one condition, and the only complication is how they combine. An intersection for a formula, an output condition for a composition. The bracket in the written answer is not decoration: it records whether the endpoint survived.

Reporting a domain as a single interval when it has a hole. For $\frac{\sqrt{x-2}}{x-5}$, writing $[2,\mathrm{\infty })$ satisfies the root and forgets the denominator. The correct answer is $[2,5)\cup (5,\mathrm{\infty })$, and the union is not a stylistic preference, $x=5$ genuinely produces division by zero.

Using the wrong bracket. $\sqrt{x-2}$ has domain $[2,\mathrm{\infty })$ with a closed bracket, since $\sqrt{0}=0$ is defined. $\ln x$ has domain $(0,\mathrm{\infty })$ with an open one, since $\ln 0$ is not. Treating the two restrictions as the same kind loses a point in one case and admits an undefined one in the other.

Keeping the direction when dividing by a negative. From , dividing by $-2$ gives , not . Testing $x=0$: $-2(0)=0$, which is not greater than 6, so 0 must be excluded, and wrongly includes it. A single test point settles the direction whenever it is in doubt.

Multiplying by an expression of unknown sign. Solving by multiplying through by $x$ assumes . For $x=-1$ the original reads , true, while the multiplied form gives , false. The correct treatment splits into cases by the sign of $x$.

Writing as a double inequality. is or , giving $(-\mathrm{\infty },1)\cup (5,\mathrm{\infty })$. Writing asserts a number is simultaneously below $-4$ and above 4. The empty set.

The companion case is different in kind: is a double inequality, , giving $(1,5)$. Checking the boundaries: at $x=1$ and $x=5$ the expression equals exactly 4, so neither endpoint belongs.

Reporting an extraneous solution. $\ln x+\ln (x-3)=\ln 10$ leads to $x^2-3x-10=0$, whose roots are $5$ and $-2$. But the original requires , and $\ln (-2)$ does not exist. Only $x=5$ is a solution, confirmed by $\ln 5+\ln 2=2.3025850930=\ln 10$.

Nothing went wrong in the algebra. Combining the logarithms enlarged the domain, so the quadratic has a root the original equation never admitted, which is why the check is part of the method.

Confusing domain with range. For $f(x)=x^2-4x+7$ the domain is all reals while the range is $[3,\mathrm{\infty })$. Answering "what values can $f$ take" with "all reals" confuses what goes in with what comes out.

$\sqrt{x-2}$ against $\ln (x-2)$.

One symbol apart in the condition, one bracket apart in the answer. The root accepts its boundary because $\sqrt{0}=0$; the logarithm rejects its own because no exponent produces 0. Treating the two restrictions as interchangeable loses or gains exactly one point.

against .

The two sets are complementary apart from the endpoints, where the expression equals exactly 4 and neither strict inequality holds. Writing the second as a double inequality yields the empty set, a difference in kind rather than in arithmetic.

$\frac{1}{\sqrt{4-x^2}}$ against $\sqrt{4-x^2}$.

The root alone has domain $[-2,2]$, closed, since $\sqrt{0}$ is fine. Put it in a denominator and the endpoints leave: $(-2,2)$. The added division converts $\ge$ into , which is the collapse two conditions undergo when one sits inside the other.

Domain against range, for $x^2-4x+7$.

Domain $(-\mathrm{\infty },\mathrm{\infty })$, since every input works. Range $[3,\mathrm{\infty })$, since not every output occurs. The two questions run in opposite directions, and the answers here share no structure: one is everything, the other is bounded below at the vertex value 3.

A solution that survives against one that does not. For $\ln x+\ln (x-3)=\ln 10$ the quadratic gives $x=5$ and $x=-2$. They are equally valid roots of the quadratic and unequally valid solutions of the equation: the domain admits the first and excludes the second. The difference is not in the algebra but in which question each root answers.

Each differs by a single feature, whether a strictness, a direction, a division or a direction of inquiry, and each difference changes the answer's shape rather than merely its value. That is why reading the formula carefully precedes solving it.

An inequality is manipulated exactly like an equation, so multiplying or dividing both sides by a negative number leaves the direction unchanged.

A function's domain is found by reading the conditions its named operations impose, so ${\displaystyle \frac{\sqrt{x-2}}{\ln (5-x)}}$ requires only $x\ge 2$ and ; that the logarithm sits in a denominator imposes nothing further.