Unit 23 · Algebra II: Polynomial, Rational and Radical Functions

Try to solve x^2 + 4 = 0. Subtract 4 from both sides: x^2 = −4. No real number works, because a real number squared is never negative. For centuries that was the end of the story. Then mathematicians invented a new number. The imaginary unit, written i, is the number whose square is −1. With it, √(−4) = √4 × √(−1) = 2i. Check: (2i)^2 = 4 × i^2 = 4 × (−1) = −4. So x = 2i and x = −2i both solve the equation.

Any square root of a negative number follows the same pattern. Pull out the −1 as i, then take the ordinary square root of what is left. √(−25) = 5i. √(−49) = 7i. √(−12) = √12 × i = 2√3 i. A number like 5i is called an imaginary number, while the numbers you already know, such as −3, 0.75 and π, are real numbers. Both kinds are needed to solve every polynomial equation.

Powers of i repeat in a cycle of four. i^1 = i, i^2 = −1, i^3 = i^2 × i = −i, and i^4 = i^2 × i^2 = (−1)(−1) = 1. After that the cycle starts over: i^5 = i, i^6 = −1. To find a large power, divide the exponent by 4 and use the remainder. For i^10, 10 = 4 × 2 + 2, so i^10 = i^2 = −1. For i^15, the remainder is 3, so i^15 = −i.

One mistake catches almost everyone. The rule √a × √b = √(ab) works only for nonnegative numbers. A student writes √(−9) × √(−4) = √36 = 6. That is wrong. Convert to i first: √(−9) = 3i and √(−4) = 2i, so the product is 3i × 2i = 6i^2 = 6 × (−1) = −6. Always rewrite negatives under a root as i before multiplying.

A complex number has the form a + bi, where a is the real part and b is the imaginary part. For 3 + 2i, the real part is 3 and the imaginary part is 2. To add or subtract, combine real parts with real parts and imaginary parts with imaginary parts. (3 + 2i) + (1 − 5i) = (3 + 1) + (2 − 5)i = 4 − 3i. Subtraction works the same way: (3 + 2i) − (1 − 5i) = 2 + 7i. Watch the sign on the second imaginary part.

To multiply, distribute every term as with binomials, then replace i^2 with −1. Take (2 + 3i)(4 − i). First: 2 × 4 = 8. Outer: 2 × (−i) = −2i. Inner: 3i × 4 = 12i. Last: 3i × (−i) = −3i^2 = −3(−1) = 3. Add them: 8 − 2i + 12i + 3 = 11 + 10i. The step most students skip is the last one, turning −3i^2 into +3. Without it the real part comes out wrong.

The conjugate of a + bi is a − bi, the same number with the sign of the imaginary part flipped. Multiply a complex number by its conjugate and the i disappears: (3 + 2i)(3 − 2i) = 9 − 6i + 6i − 4i^2 = 9 + 4 = 13. That trick lets you divide. To compute (5 + i) ÷ (2 − i), multiply top and bottom by 2 + i. Top: (5 + i)(2 + i) = 10 + 5i + 2i + i^2 = 9 + 7i. Bottom: (2 − i)(2 + i) = 4 + 1 = 5. The answer is 9/5 + (7/5)i.

A ball tossed upward from a balcony has a height that follows a quadratic. Ask when it reaches a height it never gets to, and the algebra answers with complex numbers. Take x^2 − 4x + 13 = 0. The quadratic formula says x = (−b ± √(b^2 − 4ac)) / (2a) with a = 1, b = −4, c = 13. The discriminant is b^2 − 4ac = 16 − 52 = −36. It is negative, so there are no real solutions.

Keep going anyway. √(−36) = 6i, so x = (4 ± 6i) / 2 = 2 ± 3i. The two solutions are 2 + 3i and 2 − 3i, a conjugate pair. Check one of them: (2 + 3i)^2 = 4 + 12i + 9i^2 = −5 + 12i. Then −4(2 + 3i) = −8 − 12i. Add 13: (−5 + 12i) + (−8 − 12i) + 13 = 0. It works. Divide every term by 2a, not just the real part; writing 4 ± 6i is a frequent slip.

The discriminant tells the story before you solve. Positive means two real solutions, zero means one repeated real solution, and negative means two complex conjugate solutions. For x^2 + 9 = 0 the shortcut is faster: x^2 = −9, so x = ±3i. Either way, every quadratic with real coefficients now has exactly two solutions when you count complex ones.

Dividing 4 into 93 gives 23 with 1 left over, and 93 = 4 × 23 + 1. Polynomials divide the same way. The polynomial being divided is the dividend, the one you divide by is the divisor, the result is the quotient, and what is left is the remainder. Divide 2x^3 + 3x^2 − 5x + 4 by x + 2. Ask how many times x goes into 2x^3: the answer is 2x^2, the first term of the quotient.

Multiply 2x^2 by the divisor: 2x^2(x + 2) = 2x^3 + 4x^2. Subtract from the dividend: (2x^3 + 3x^2) − (2x^3 + 4x^2) = −x^2. Bring down the −5x to get −x^2 − 5x. Repeat: −x^2 ÷ x = −x. Multiply: −x(x + 2) = −x^2 − 2x. Subtract: −5x − (−2x) = −3x. Bring down the 4: −3x + 4. Once more: −3x ÷ x = −3, and −3(x + 2) = −3x − 6. Subtract: 4 − (−6) = 10.

The quotient is 2x^2 − x − 3 and the remainder is 10. In one line: 2x^3 + 3x^2 − 5x + 4 = (x + 2)(2x^2 − x − 3) + 10. Check by multiplying. (x + 2)(2x^2 − x − 3) = 2x^3 − x^2 − 3x + 4x^2 − 2x − 6. That simplifies to 2x^3 + 3x^2 − 5x − 6. Adding 10 gives the original polynomial. Sign errors in the subtract step cause most wrong answers. Subtract the whole line, including the minus signs.

If a power is missing, hold its place with a zero. To divide x^3 − 8 by x − 2, write the dividend as x^3 + 0x^2 + 0x − 8 so the columns line up. The cycle gives x^2, then 2x, then 4, with remainder 0. So x^3 − 8 = (x − 2)(x^2 + 2x + 4). A remainder of 0 means the divisor is a factor, an idea the next two lessons build on.

When the divisor is x − c, a shortcut called synthetic division uses only the coefficients. Divide p(x) = x^3 − 4x^2 + x + 6 by x − 3, so c = 3. Write the coefficients 1, −4, 1, 6 in a row. Bring down the 1. Multiply by 3 and add to the next coefficient: 1 × 3 = 3, and −4 + 3 = −1. Multiply again: −1 × 3 = −3, and 1 + (−3) = −2. Once more: −2 × 3 = −6, and 6 + (−6) = 0.

The bottom row reads 1, −1, −2, 0. The last number is the remainder, 0. The others are the quotient’s coefficients, one degree lower than the dividend: x^2 − x − 2. So x^3 − 4x^2 + x + 6 = (x − 3)(x^2 − x − 2). Two cautions. Use c, not −c: dividing by x + 1 means c = −1. And add at each step, never subtract; the subtracting is built into the sign of c.

The remainder theorem explains the last number. When a polynomial p(x) is divided by x − c, the remainder is exactly p(c). Check with the example: p(3) = 27 − 36 + 3 + 6 = 0, matching the remainder 0. This gives a fast way to evaluate a polynomial, meaning to find its value at a number. For q(x) = 2x^3 − 5x + 1 at x = 2, synthetic division with coefficients 2, 0, −5, 1 gives 2, 4, 3, 7, so q(2) = 7. Direct check: 16 − 10 + 1 = 7.

The remainder theorem has a powerful special case. If p(c) = 0, the remainder is 0, so x − c divides p(x) evenly and is a factor. The reverse holds too. This is the factor theorem: x − c is a factor of p(x) if and only if p(c) = 0. A number c with p(c) = 0 is called a zero of the function. Finding zeros and finding factors are the same job.

Factor p(x) = x^3 − 7x + 6. Try x = 1: p(1) = 1 − 7 + 6 = 0, so x − 1 is a factor. Synthetic division with coefficients 1, 0, −7, 6 and c = 1 gives 1, 1, −6, 0. The quotient is x^2 + x − 6, which factors as (x + 3)(x − 2). So p(x) = (x − 1)(x − 2)(x + 3), and the zeros are 1, 2 and −3. Check the last one: p(−3) = −27 + 21 + 6 = 0.

Which values should you try? For integer coefficients, any rational root p/q must have p dividing the constant term and q dividing the leading coefficient. For 2x^3 − 3x^2 − 3x + 2 the constant is 2 and the leading coefficient is 2, so the candidates are ±1, ±2 and ±1/2. Test x = 2: 16 − 12 − 6 + 2 = 0. Synthetic division gives 2, 1, −1, 0, so the quotient is 2x^2 + x − 1 = (2x − 1)(x + 1). The zeros are 2, 1/2 and −1.

The candidate list is only a list of suspects. Many will fail, and a polynomial can have no rational zeros at all. Test the small integers first, and stop dividing once the quotient is a quadratic you can factor or solve with the formula. A common error is testing x = 2 by plugging in x − 2; the test value is c, the number that makes the factor zero.

A linear equation has one solution, a quadratic has two, and the pattern continues. The fundamental theorem of algebra states that every polynomial of degree n ≥ 1 has exactly n zeros, as long as you count complex zeros and count a repeated zero as many times as it repeats. Carl Friedrich Gauss gave a proof in his 1799 doctoral thesis. The theorem does not find the zeros for you, but it tells you when to stop looking.

Take p(x) = x^3 − x^2 + 4x − 4. Its degree is 3, so expect three zeros. Factor by grouping: x^2(x − 1) + 4(x − 1) = (x − 1)(x^2 + 4). The factor x − 1 gives the zero 1. The factor x^2 + 4 gives x^2 = −4, so x = 2i and x = −2i. Three zeros: 1, 2i, −2i. Only one shows up as an x-intercept on the graph, because complex zeros have no place on a real number line.

When a polynomial has real coefficients, complex zeros arrive in conjugate pairs. If 1 + i is a zero, so is 1 − i. That lets you rebuild a polynomial from its zeros. For zeros 2, 1 + i and 1 − i: the pair gives (x − (1 + i))(x − (1 − i)) = (x − 1)^2 − i^2 = x^2 − 2x + 1 + 1 = x^2 − 2x + 2. Multiply by (x − 2): x^3 − 2x^2 + 2x − 2x^2 + 4x − 4 = x^3 − 4x^2 + 6x − 4.

Write a polynomial in factored form and its graph is almost drawn for you. Take f(x) = (x + 1)(x − 2)^2. The zeros are −1 and 2, so the graph has x-intercepts at those points. But the factor x − 2 appears twice. The number of times a factor repeats is the zero’s multiplicity. Here −1 has multiplicity 1 and 2 has multiplicity 2. The degree, found by adding the multiplicities, is 1 + 2 = 3.

Multiplicity changes how the graph meets the axis. Test points on each side of x = 2. f(1) = (2)(−1)^2 = 2, which is positive. f(3) = (4)(1)^2 = 4, also positive. The graph does not change sign at 2, so it touches the axis there and bounces back up. Now test around x = −1. f(−2) = (−1)(−4)^2 = −16, negative, while f(0) = (1)(4) = 4, positive. The sign flips, so the graph crosses at −1.

The rule: odd multiplicity means the graph crosses, and even multiplicity means it touches and turns. A multiplicity of 3 crosses but flattens out at the intercept, like the graph of x^3 near the origin. The y-intercept is f(0), which for our example is 4. Between the intercepts the graph has turning points, hills and valleys; a degree-3 graph has at most 2 of them.

Zoom far out on the graph of f(x) = −2x^3 + 5x^2 + 100 and the bumps disappear. What remains is the shape of its leading term, −2x^3, the term with the highest power. At x = 100, the terms are −2,000,000, then 50,000, then 100. The leading term dwarfs the rest, and f(100) = −1,949,900. The direction of the graph’s two ends, called its end behavior, depends only on that leading term.

Two facts decide it. The degree tells whether the ends match. Even degree means both ends go the same way, like a parabola. Odd degree means the ends go opposite ways, like a line. The sign of the leading coefficient tells which way. For positive leading coefficient, the right end rises. For negative, the right end falls. So −2x^3 has odd degree and a negative coefficient: the right end goes down and the left end goes up.

Practice on f(x) = 3x^4 − 7x^3 + x − 9. The leading term is 3x^4: even degree, positive coefficient, so both ends rise. Notice that the term 50x in 7x^2 − 3x^6 + 50x does not lead, even though it is written last; the leading term is −3x^6, so both ends fall. Between the ends a degree-n graph has at most n − 1 turning points, which limits how wiggly it can be.

Return to the 20-inch square sheet with a square of side x cut from each corner. Folding up the sides makes a box with height x. Each side of the base loses x from both ends, so the base is 20 − 2x by 20 − 2x. The volume is V(x) = x(20 − 2x)^2. This equation is a model: a function that stands in for the real situation. Multiplied out it is a cubic function, 4x^3 − 80x^2 + 400x.

Not every x makes sense. The cut must be positive, and two cuts cannot use up the whole 20 inches, so 0 < x < 10. That interval is the model’s domain. Now test values. V(1) = 1 × 18^2 = 324. V(2) = 2 × 16^2 = 512. V(3) = 3 × 14^2 = 588. V(4) = 4 × 12^2 = 576. V(5) = 5 × 10^2 = 500. The volume climbs, peaks between 3 and 4, and falls.

To pin down the maximum, test between 3 and 4. V(3.3) = 3.3 × 13.4^2 = 3.3 × 179.56 = 592.5. V(3.5) = 3.5 × 13^2 = 591.5. The peak is near x = 3.3 inches, giving about 593 cubic inches. A graphing tool shows the same hill. Students sometimes report x = 3.3 as the answer to the question about volume; read the question again, because the height and the volume are different answers.

Rectangular sheets follow the same recipe with two base lengths. For a 16-inch by 10-inch sheet, V(x) = x(16 − 2x)(10 − 2x), and the domain is 0 < x < 5 because the short side runs out first. V(2) = 2 × 12 × 6 = 144 cubic inches. Whatever the numbers, write each dimension in terms of x, multiply, then explore the domain.

A drone flies along a straight path while a ball arcs through the air. Where could they meet? That is a system of equations with one linear and one quadratic equation. Take y = x^2 − 3 and y = 2x. Both equal y, so set them equal: x^2 − 3 = 2x. Move everything to one side: x^2 − 2x − 3 = 0. Factor: (x − 3)(x + 1) = 0, so x = 3 or x = −1.

Each x needs its y. Use the line: when x = 3, y = 6; when x = −1, y = −2. The intersection points are (3, 6) and (−1, −2). Check them in the parabola: 3^2 − 3 = 6 and (−1)^2 − 3 = −2. Both work. A line and a parabola can meet at two points, one point if the line just grazes the curve, or nowhere at all. If the quadratic has a negative discriminant, there are no real intersections.

A circle works the same way with substitution. Solve x^2 + y^2 = 25 together with y = x + 1. Replace y: x^2 + (x + 1)^2 = 25, so 2x^2 + 2x + 1 = 25. Then 2x^2 + 2x − 24 = 0, and dividing by 2 gives x^2 + x − 12 = 0 = (x + 4)(x − 3). So x = 3 with y = 4, or x = −4 with y = −3. Check: 9 + 16 = 25 and 16 + 9 = 25. Forgetting to find the y-values is the most common way to lose the answer.

Splitting a $60 bill among x friends costs each person 60/x dollars. That formula makes no sense when x = 0, because nobody can divide by zero. A rational expression is a fraction whose numerator and denominator are polynomials, like (x^2 − 9)/(x^2 − x − 6). Its domain is every real number except the ones that make the denominator zero. Those are the excluded values.

Find the excluded values by factoring the denominator and setting each factor to zero. For x^2 − x − 6 = (x − 3)(x + 2), the excluded values are x = 3 and x = −2. Always find them from the original denominator, before any canceling, because canceling can hide a value that is still not allowed.

To simplify, factor the numerator too and cancel factors shared by top and bottom. (x^2 − 9)/(x^2 − x − 6) = (x − 3)(x + 3) / ((x − 3)(x + 2)). The factor x − 3 cancels, leaving (x + 3)/(x + 2), with x ≠ 3 and x ≠ −2 still attached. The expressions agree everywhere in the domain, but the simplified one no longer shows the x = 3 restriction on its own.

The classic mistake is canceling terms instead of factors. A student sees (x + 3)/(x + 2) and crosses out the x’s to get 3/2. Test it: at x = 1, (1 + 3)/(1 + 2) = 4/3, not 3/2. You can only cancel something that multiplies the entire numerator and the entire denominator. When in doubt, plug in a number and compare.

Multiplying rational expressions is like multiplying 2/3 × 9/4: factor, cancel, then multiply what is left. Take (x^2 − 1)/(x + 2) × (x + 2)/(x − 1). Factor first: (x − 1)(x + 1)/(x + 2) × (x + 2)/(x − 1). Now cancel any factor that appears in a numerator and a denominator, even across the two fractions. The x + 2 pair cancels and the x − 1 pair cancels, leaving x + 1. Excluded values: x ≠ −2 and x ≠ 1.

To divide, multiply by the reciprocal of the second fraction, the fraction flipped over. Compute x/(x + 3) ÷ x^2/(x^2 − 9). Flip the second: x/(x + 3) × (x^2 − 9)/x^2. Factor: x/(x + 3) × (x − 3)(x + 3)/(x × x). Cancel one x and the x + 3: the result is (x − 3)/x. Flip only the divisor. Flipping the first fraction gives the reciprocal of the right answer.

Another trap is canceling before factoring. In (3x)/(x − 4) × (x^2 − 16)/(6x^2), nothing cancels until x^2 − 16 becomes (x − 4)(x + 4). Then x − 4 cancels, 3x against 6x^2 leaves 1 against 2x, and the product is (x + 4)/(2x). Remember that every value that was excluded in the original problem, including any zeros of the divisor’s numerator, stays excluded.

Adding 1/4 + 1/6 needs a common denominator, 12: 3/12 + 2/12 = 5/12. Rational expressions work the same way. To add 3/x + 2/(x + 1), the least common denominator is x(x + 1), the smallest expression both denominators divide. Rewrite each fraction: 3/x = 3(x + 1)/(x(x + 1)) and 2/(x + 1) = 2x/(x(x + 1)). Add the numerators: 3(x + 1) + 2x = 3x + 3 + 2x = 5x + 3. The sum is (5x + 3)/(x(x + 1)).

When denominators share a factor, do not just multiply them. For 1/(x^2 − 4) + 1/(x + 2), factor x^2 − 4 = (x − 2)(x + 2). The LCD is (x − 2)(x + 2), not (x^2 − 4)(x + 2). Only the second fraction needs adjusting: 1/(x + 2) = (x − 2)/((x − 2)(x + 2)). Adding gives (1 + x − 2)/((x − 2)(x + 2)) = (x − 1)/(x^2 − 4).

Subtraction adds one hazard: the minus sign applies to the whole second numerator. Compute 1/(x − 2) − 1/(x + 2). Over the LCD (x − 2)(x + 2), the numerators are (x + 2) − (x − 2) = x + 2 − x + 2 = 4. The result is 4/(x^2 − 4). Dropping the parentheses gives x + 2 − x − 2 = 0, which is wrong. And never add denominators: 1/x + 1/x is 2/x, not 2/(2x).

Look at f(x) = 1/(x − 2) near x = 2. At x = 2.1 the value is 1/0.1 = 10. At 2.01 it is 100, and at 2.001 it is 1,000. From the other side, f(1.9) = −10 and f(1.99) = −100. The graph shoots up on the right of 2 and down on the left, hugging the line x = 2 without touching it. That line is a vertical asymptote. The function is undefined at x = 2 itself.

Now take g(x) = (x^2 − 1)/(x^2 − 3x + 2). The denominator is zero at x = 1 and x = 2, but the two spots behave differently. Factor: (x − 1)(x + 1)/((x − 1)(x − 2)). The factor x − 1 cancels, leaving (x + 1)/(x − 2). The factor x − 2 does not cancel, so x = 2 is a vertical asymptote. At x = 1 the simplified form gives (1 + 1)/(1 − 2) = −2, so the graph is a smooth curve with one missing point at (1, −2). That missing point is a hole.

The rule: factor top and bottom. A denominator factor that cancels makes a hole; a denominator factor that survives makes a vertical asymptote. Both are excluded from the domain. A common error is to cancel first and forget the hole, reporting the domain as everything except 2. The original denominator was zero at 1 too, so 1 stays excluded.

What does f(x) = (3x^2 + 1)/(x^2 − 4) do when x is huge? At x = 100 the value is 30,001/9,996, which is about 3.001. At x = 1,000 it is even closer to 3. Far to the right and far to the left, the graph flattens toward the line y = 3. That line is a horizontal asymptote. It describes end behavior, the same idea as for polynomials, but now the ends level off instead of shooting away.

For large x only the leading terms matter. So compare the degree of the numerator with the degree of the denominator. Case one, top degree lower: (x + 1)/(x^2 + 1) behaves like x/x^2 = 1/x. That shrinks toward 0, so the asymptote is y = 0. Case two, degrees equal: (3x^2 + 1)/(x^2 − 4) behaves like 3x^2/x^2 = 3. The asymptote is y = 3, the ratio of the leading coefficients.

Case three, top degree higher: x^2/(x − 1) behaves like x^2/x = x, which keeps growing. There is no horizontal asymptote. Unlike a vertical asymptote, a horizontal one can be crossed in the middle of the graph; it only governs the far ends. Do not confuse the two kinds: vertical asymptotes come from the denominator’s zeros, horizontal ones from comparing degrees.

A 240-mile drive takes 4 hours at 60 miles per hour but 6 hours at 40 miles per hour. Faster speed, less time. The product is always the same: 60 × 4 = 240 and 40 × 6 = 240. When two quantities have a constant product, they show inverse variation. In symbols, t = 240/v, or in general y = k/x, where k is the constant of variation. Compare that with direct variation, y = kx, where the two quantities rise together.

To solve an inverse variation problem, find k first. Suppose y varies inversely with x, and y = 6 when x = 4. Then k = xy = 24, so the rule is y = 24/x. When x = 8, y = 24/8 = 3. Doubling x from 4 to 8 halved y from 6 to 3. A frequent error is to assume that doubling x doubles y; that is direct variation, and the numbers in the problem will show which kind you have.

Work crews follow the same pattern. If 12 workers finish a job in 10 hours, the job takes 12 × 10 = 120 worker-hours. With 8 workers, the time is 120/8 = 15 hours. Boyle’s law in chemistry is another example: at a fixed temperature, the pressure of a gas times its volume stays constant, so squeezing a gas to half its volume doubles its pressure. The graph of y = k/x is a curve with both axes as asymptotes.

A rational equation has a variable in at least one denominator. Solve 3/x + 1/2 = 2. The least common denominator of x and 2 is 2x. Multiply every term by 2x, a step called clearing denominators: 2x × 3/x = 6, 2x × 1/2 = x, and 2x × 2 = 4x. The equation becomes 6 + x = 4x, so 6 = 3x and x = 2. Check in the original: 3/2 + 1/2 = 2. It works.

Sometimes the algebra produces an answer that the original equation cannot accept. Solve x/(x − 3) = 3/(x − 3) + 2. Multiply every term by x − 3: x = 3 + 2(x − 3) = 3 + 2x − 6 = 2x − 3. So x = 3. But x = 3 makes the original denominators zero, so it is not allowed. It is an extraneous solution, an answer created by the multiplication step. This equation has no solution.

The safe habit is to write down the excluded values before you start. For x/(x − 3), that is x ≠ 3. Solve, then compare each answer with that list. Anything on the list is thrown out. Also multiply every term, including the constants; forgetting to multiply the 2 by x − 3 in the example above is the most common slip.

Two hoses fill a pool. Alone, the first takes 6 hours and the second takes 3 hours. Together? Think in work rates: the first hose fills 1/6 of the pool per hour and the second fills 1/3 per hour. Together they fill 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2 of the pool per hour, so the job takes 2 hours. In general 1/a + 1/b = 1/t. Averaging the times, (6 + 3)/2 = 4.5, is wrong; two hoses must be faster than the faster hose alone.

Round trips use the same logic with distance and time. Drive 120 miles out at 60 miles per hour and back at 40. Time out: 120/60 = 2 hours. Time back: 120/40 = 3 hours. Average speed is total distance over total time: 240/5 = 48 miles per hour. The general formula for speeds a and b is 2ab/(a + b), which for 60 and 40 gives 4,800/100 = 48. This kind of average is called the harmonic mean.

Now run it backward. You drive out at an unknown speed x, return at 30 miles per hour, and the average speed is 40. Set up 2 × x × 30/(x + 30) = 40. Multiply by x + 30: 60x = 40x + 1,200, so 20x = 1,200 and x = 60. Check: out at 60 and back at 30 over 60 miles takes 1 + 2 = 3 hours for 120 miles, and 120/3 = 40. The slower leg always takes more time, which is why the average leans toward it.

The time for a pendulum to swing, the speed of a wave, the distance to the horizon: many formulas put a variable under a radical, a root sign. The function f(x) = √(x − 3) is a square root function. Real square roots need nonnegative inputs, so x − 3 ≥ 0 and the domain is x ≥ 3. Its graph starts at (3, 0) and rises slowly: f(4) = 1, f(7) = 2, f(12) = 3. A cube root like ∛x has no such limit, because negative numbers have cube roots.

To solve √(2x + 1) = 5, isolate the root, then square both sides to remove it. That gives 2x + 1 = 25, so 2x = 24 and x = 12. Check in the original: √(24 + 1) = √25 = 5. If the equation has extra terms, move them first. √(x − 1) + 4 = 7 becomes √(x − 1) = 3. Squaring gives x − 1 = 9, so x = 10. For a cube root, cube both sides instead. ∛(x − 1) = 2 gives x − 1 = 8, so x = 9.

Squaring can invent solutions. Solve √(x + 6) = x. Square: x + 6 = x^2, so x^2 − x − 6 = 0 and (x − 3)(x + 2) = 0. Candidates: x = 3 and x = −2. Check x = 3: √9 = 3, correct. Check x = −2: √4 = 2, but the right side is −2, so it fails. Only x = 3 works; −2 is extraneous. A square root is never negative, so any candidate that makes the root equal a negative number must be thrown out. Always check.

Converting Celsius to Fahrenheit uses F = 9C/5 + 32. Converting back is the same relationship run in reverse: C = 5(F − 32)/9. Each formula undoes the other. Check: 20°C gives F = 36 + 32 = 68, and 68°F gives C = 5 × 36/9 = 20. Two functions that undo each other are inverse functions, written f and f^−1.

To find an inverse from a formula, swap x and y, then solve for y. For f(x) = 2x + 3, write y = 2x + 3, swap to get x = 2y + 3, and solve: y = (x − 3)/2. So f^−1(x) = (x − 3)/2. Test with a number: f(5) = 13, and f^−1(13) = (13 − 3)/2 = 5. The graph of f^−1 is the graph of f reflected across the line y = x.

Not every function has an inverse. f(x) = x^2 sends both 2 and −2 to 4, so running it backward from 4 gives two answers, and an inverse must give one. A function has an inverse only if it is one-to-one: different inputs always give different outputs. The horizontal line test checks this on a graph: if any horizontal line crosses the graph twice, there is no inverse. Restricting x^2 to x ≥ 0 fixes it, and the inverse is √x. Likewise x^3 has the inverse ∛x for all x.

A jacket is on sale for 20% off, and then sales tax of 10% is added; treat that tax rate as an example figure. Two functions act in turn. The discount is d(p) = 0.8p and the tax is t(p) = 1.1p. On a $50 jacket, first d(50) = 40, then t(40) = 44. Feeding one output into another function is composition, written t(d(p)). The inner function d acts first, then the outer function t.

Composition with formulas works by substitution. Let f(x) = x + 2 and g(x) = x^2. Then f(g(3)) = f(9) = 11, and g(f(3)) = g(5) = 25. In general f(g(x)) = x^2 + 2, while g(f(x)) = (x + 2)^2. The two are different functions, so order matters: adding 2 then squaring is not the same as squaring then adding 2. Read f(g(x)) from the inside out.

Composition is the test for inverses. Two functions are inverses exactly when f(g(x)) = x and g(f(x)) = x. Try f(x) = 2x + 3 and g(x) = (x − 3)/2. Compute f(g(x)) = 2 × (x − 3)/2 + 3 = x − 3 + 3 = x. And g(f(x)) = (2x + 3 − 3)/2 = 2x/2 = x. Each undoes the other, so they are inverses. If either composition gives something other than x, the two functions are not inverses.