Functions, Domains and Inequalities

$\frac{1}{x}$ tells you what to do with a number. It does not tell you which numbers are allowed, and that answer is not a matter of choice: division by zero is undefined, so 0 cannot be an input, and the set of permitted inputs is forced by the arithmetic itself.

That set is the domain, and it is part of the function rather than a detail attached afterwards. Two functions with the same formula and different domains are different functions, which is why $\sqrt{x}$ on $[0,\mathrm{\infty })$ and $\sqrt{x}$ on $[4,9]$ answer different questions.

Finding a domain is a search for failure. Only three operations can fail on real numbers: division at a zero denominator, the square root on a negative, the logarithm at zero and below. Everything else, including polynomials, sums, products and odd roots, accepts every real input. So reading a formula for its domain means scanning for those three and writing one condition each.

The conditions then have to hold together. For $\frac{\sqrt{x-2}}{x-5}$ the root requires $x\ge 2$ and the denominator requires $x\ne 5$, giving $[2,5)\cup (5,\mathrm{\infty })$, a set with a hole, which is what interval notation exists to write down.

This matters beyond bookkeeping. Every question the calculus units ask (does this limit exist, is this differentiable, can this be integrated) is a question about a point in the domain, and asking it at a point outside is a category error rather than a hard problem.

Why only three operations restrict. Addition, subtraction and multiplication are defined on all of $\mathbb{R}$, so any polynomial is total. Division fails only at zero, since no real number times 0 gives a nonzero result. The square root fails on negatives because no real number squares to a negative, and the odd root does not fail, since $(-2{)}^3=-8$ gives $\sqrt[3]{-8}=-2$. The logarithm fails at zero and below because for every real $y$, so no exponent produces a non-positive value.

Each restriction traces to a fact about the real numbers rather than to a convention, which is why the list is exactly three items long and does not grow.

Why the conditions intersect rather than accumulate. Each operation must succeed for the formula to have a value, so an input is admissible only if it satisfies every condition. For $\frac{\sqrt{x-2}}{x-5}$, $x=1$ fails the root and $x=5$ fails the division; both are excluded, and the domain is the intersection $\{x\ge 2\}\cap \{x\ne 5\}$.

Why $\mathrm{\infty }$ always takes a parenthesis. The bracket means the endpoint is included, and $\mathrm{\infty }$ is not a number that could be. Writing $[2,\mathrm{\infty }]$ claims a largest real number exists.

Why a composition's domain is not an intersection. $(g\circ f)$ needs $x$ in $f$'s domain and $f(x)$ in $g$'s domain. A condition on the output, not on $x$ directly. With $f(x)=x-1$, total, and $g(u)=\sqrt{u}$, the composite $\sqrt{x-1}$ requires $x\ge 1$, a restriction neither domain contains on its own.

Why a negative multiplier reverses an inequality. Multiplying by $-1$ reflects the number line through 0, and reflection exchanges left and right. From , multiplying by $-1$ gives , which is visibly correct. An equation has no direction to reverse, which is why the rule appears here and not there.

Multiplying by an expression whose sign is unknown, such as $x$, is therefore invalid without splitting into cases, and that is why is not solved by multiplying through by $x$.

Why and unfold differently. $|u|$ is the distance from $u$ to 0. Being within $c$ of 0 confines $u$ to one interval, . Being further than $c$ from 0 admits two rays, or , which cannot be written as a single double inequality, attempting describes the empty set.

Why extraneous solutions arise from valid steps. $\ln x+\ln (x-3)$ requires . Rewriting it as $\ln (x(x-3))$ requires only , which also holds for . The step is a correct identity where both sides are defined, and the rewritten form is defined more widely, so the new equation can have solutions the original does not. Checking candidates against the original domain is the method's last step rather than an optional caution.

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.

To find a natural domain.

1. Scan for the three restricting operations. Division, even roots, logarithms. Nothing else in an elementary formula can fail.
2. Write one condition per occurrence.

1. Solve each condition for $x$ separately.
2. Intersect them. An input must satisfy every condition at once.
3. Write the result in interval notation, with a union where the set has gaps.

Watch for conditions that collapse. A root inside a denominator, $\frac{1}{\sqrt{g}}$, needs $g\ge 0$ and $g\ne 0$ together, which is simply .

To find the domain of a composition $g(f(x))$. Require $x$ in $f$'s domain and $f(x)$ in $g$'s domain. The second is a condition on the output, so solve $f(x)\in \mathrm{dom}g$ as an inequality in $x$. The result may be smaller than either domain.

To solve a linear inequality. Proceed as for an equation, with one exception: reverse the direction whenever multiplying or dividing by a negative number. Never multiply by an expression whose sign is unknown, split into cases instead.

To solve an absolute-value inequality. The two forms unfold differently:

• becomes the double inequality . One interval.
• becomes or . A union of two rays.

If , the first has no solutions and the second has all of them, since $|u|\ge 0$ always.

To solve an exponential equation. Write both sides as powers of a common base and equate the exponents: $3^{2x}=81=3^4$ gives $2x=4$. Where no common base is available, apply a logarithm to both sides.

To solve a logarithmic equation.

1. State the domain first, from the arguments of every logarithm.
2. Combine logarithms using $\log u+\log v=\log (uv)$.
3. Convert to exponential form and solve.
4. Check every candidate against the domain from step 1, discarding those outside it.

Skipping step 4 reports a domain the formula does not have. Combining logarithms enlarges the domain, so the algebra can produce roots the original equation never admitted.

Checks. Substitute a test point from each interval of the claimed solution set, and one from outside it. The first should satisfy the inequality and the second should not. For a domain, try a value just inside and just outside each endpoint; the second should fail the arithmetic rather than merely look wrong.

1. A domain with a gap: $f(x)={\displaystyle \frac{\sqrt{x-2}}{x-5}}$.

Two restricting operations, so two conditions:

• the square root needs $x-2\ge 0$, so $x\ge 2$;
• the denominator needs $x-5\ne 0$, so $x\ne 5$.

Both must hold, giving

Check by testing: $x=1.9$ fails the root; $x=2$ is admitted, with $f(2)=0$; $x=4.9$ is admitted; $x=5$ divides by zero; $x=5.1$ is admitted.

The closed bracket at 2 is correct because $\sqrt{0}=0$ is defined; the parentheses at 5 mark a genuine exclusion.

2. A collapsing pair of conditions: $g(x)={\displaystyle \frac{1}{\sqrt{4-x^2}}}$.

The root needs $4-x^2\ge 0$; the denominator needs $\sqrt{4-x^2}\ne 0$, which means $4-x^2\ne 0$. Together these are the single strict condition , that is :

Check: at $x=2.5$, $3$ and $5$ the computation fails, since $4-x^2$ is negative. Both endpoints are excluded even though the root alone would have admitted them.

3. Domain and range are different questions: $h(x)=x^2-4x+7$.

Nothing can fail, so $\mathrm{dom}h=(-\mathrm{\infty },\mathrm{\infty })$.

For the range, complete the square:

Since $(x-2{)}^2\ge 0$, the smallest value is 3, attained at $x=2$, and every larger value is attained. So the range is $[3,\mathrm{\infty })$.

Check: $h$ at $0,1,2,3,4$ gives $7,4,3,4,7$, and a scan of 200,000 random inputs produced a minimum of $3.0000005$, never below 3.

4. An absolute-value inequality: .

The form is , so it unfolds to a double inequality:

Add 6 throughout: . Divide by 2, positive, so no reversal: , that is $(1,5)$.

Check at the boundaries: at $x=1$ and $x=5$ the expression equals exactly 4, which is not less than 4, so neither endpoint belongs. At $x=3$ it is 0, and at $x=0.9$ and $x=5.1$ it is $4.2$, outside, as required.

Had the inequality been , the answer would be the union $(-\mathrm{\infty },1)\cup (5,\mathrm{\infty })$ instead.

5. A logarithmic equation with an extraneous root: $\ln x+\ln (x-3)=\ln 10$.

Domain first. $\ln x$ needs and $\ln (x-3)$ needs . Together: .

Combine and solve.

which factors as $(x-5)(x+2)=0$, giving $x=5$ and $x=-2$.

Check against the domain. . $x=-2$ is outside, $\ln (-2)$ does not exist, so it is discarded.

Verify the survivor: $\ln 5+\ln 2=2.3025850930$ and $\ln 10=2.3025850930$.

The algebra was correct throughout. Combining the logarithms widened the domain, so the quadratic acquired a root the original equation never had.

An inequality is manipulated like an equation with one exception, and the exception is the source of most wrong answers in this material.

The rule. Multiplying or dividing both sides by a negative number reverses the direction.

Why. Multiplying by $-1$ reflects the number line through 0, and reflection exchanges left and right. From , multiplying by $-1$ gives , which is plainly right, since $-2$ lies to the right of $-5$. An equation has no direction to preserve, which is why the rule appears only here.

The check that costs nothing. Substitute a value that the claimed answer admits. For with the wrong answer , test $x=0$: the left side is $0$, which is not greater than 6, so $0$ cannot belong, and wrongly includes it. Test $x=-4$ against the correct answer: .

One test point settles the direction whenever it is in doubt, and it takes a few seconds.

The harder case: an unknown sign. Solving by multiplying through by $x$ assumes without saying so. Checking $x=-1$: the original reads , true, while the multiplied form gives , false. The two are not equivalent.

The sound treatment splits into cases. For , multiplying preserves the direction and gives , so . For , the direction reverses and gives , which every negative $x$ satisfies. The solution is therefore $(-\mathrm{\infty },0)\cup (\frac{1}{2},\mathrm{\infty })$. A union that the single-step method misses entirely.

A related error: squaring. Squaring both sides is not reversible either, since does not imply when signs differ: but . Squaring is safe only when both sides are known non-negative, and otherwise introduces solutions that must be checked and discarded.

The habit worth forming. After solving any inequality, pick one number from inside the claimed solution set and one from outside, and test both against the original. The first should satisfy it and the second should not. That single check catches reversed directions, lost cases and extraneous roots alike.