Functions: powers, roots, inverses and polynomials

Answer: $(-3, -4)$

$y = \dfrac{1}{x}$ has the asymptotes $x = 0$ and $y = 0\mathclose{,}$ crossing at the origin. The shifts move every point of the picture, the asymptotes included: the vertical one to $x = -3$ and the horizontal one to $y = -4\mathclose{.}$ They cross at $(-3, -4)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Shifted the vertical asymptote the wrong way → $(3, -4)$
• Kept the asymptotes of $y = \dfrac{1}{x}$ → $(0, 0)$

Answer: $\dfrac{8 - 7 x}{3 x - 1}$

Write $y = \dfrac{x + 8}{3 x + 7}$ and solve for $x\mathclose{:}$ $y\left(3 x + 7\right) = x + 8\mathclose{,}$ so $3 x y + 7 y = x + 8\mathclose{.}$ Gather the $x$ terms: $x\left(3 y - 1\right) = 8 - 7 y\mathclose{,}$ so $x = \dfrac{8 - 7 y}{3 y - 1}\mathclose{.}$ Swap the letters: $f^{-1}(x) = \dfrac{8 - 7 x}{3 x - 1}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the reciprocal $\frac{1}{f(x)}$ instead of the inverse → $\frac{3 x + 7}{x + 8}$
• Swapped $a$ and $d$ but did not change the signs of $b$ and $c$ → $\frac{7 x + 8}{3 x + 1}$

Answer: $2$

A cube is undone by a cube root, and unlike a square root it keeps the sign: $\sqrt[3]{-8} = -2\mathclose{.}$ So $x - 4 = -2$ and $x = 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided by 3 instead of taking the cube root → $\frac{4}{3}$
• Took the cube root but lost the minus sign → $6$

Answer: $x = 7$

Square both sides: $x + 74 = \left(x + 2\right)^{2}\mathclose{,}$ so $x^{2} + 3 x - 70 = 0$ and $x = 7$ or $x = -10\mathclose{.}$ SQUARING CAN ADD SOLUTIONS, so check both in the original: at $x = 7$ both sides are $9\mathclose{;}$ at $x = -10$ the right-hand side is $-8\mathclose{,}$ negative, while a square root never is. Only $x = 7$ survives.

Common mistakes, and the answer each one gives:

• Kept both solutions of the squared equation without checking them → $x = -10 \text{ or } x = 7$
• Gave the solution that fails the check → $x = -10$

Answer: $15$

The highest power in the product comes from the highest powers of the factors: $\left(x^{2}\right)^{3} \cdot \left(2x^{3}\right)^{3}$ has degree $6 + 9 = 15\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added only the exponents outside the brackets → $6$
• Multiplied the degrees instead of adding them → $54$

Answer: $x = -3,\ x = 0 \text{ or } x = 1$

Every term has a factor $x\mathclose{,}$ so take it out: $x\left(x^{2} + 2 x - 3\right) = 0\mathclose{.}$ Either $x = 0\mathclose{,}$ or $x^{2} + 2 x - 3 = \left(x - 1\right)\left(x + 3\right) = 0\mathclose{.}$ The solutions are $x = -3,\ x = 0 \text{ or } x = 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided both sides by $x$ and lost the solution $x = 0$ → $x = -3 \text{ or } x = 1$
• Read the solutions off the brackets with the signs unchanged → $x = -1,\ x = 0 \text{ or } x = 3$

Answer: $y = 2\left(x - 3\right)^{4} + 1$

In that order: the stretch gives $y = 2x^{4}\mathclose{,}$ and the shifts then give $y = 2\left(x - 3\right)^{4} + 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Stretched AFTER shifting up or down, so the vertical shift was stretched too → $y = 2\left(\left(x - 3\right)^{4} + 1\right)$
• Shifted the wrong way horizontally → $y = 2\left(x + 3\right)^{4} + 1$

Answer: $x \le -1 \text{ or } 1 \le x \le 5$

Factorise: $x^{3} - 5 x^{2} - x + 5 = \left(x + 1\right)\left(x - 1\right)\left(x - 5\right)\mathclose{,}$ which is zero at $x = -1\mathclose{,}$ $1$ and $5\mathclose{.}$ Far to the right the sign is that of the leading term, positive, and it changes at each of the three roots. So the polynomial is positive for $-1 < x < 1 \text{ or } x > 5$ and negative for $x < -1 \text{ or } 1 < x < 5\mathclose{.}$ The answer is $x \le -1 \text{ or } 1 \le x \le 5\mathclose{,}$ with the roots themselves included.

Common mistakes, and the answer each one gives:

• Treated it like a quadratic, with only the outer two roots → $-1 \le x \le 5$
• Took the sign chart the wrong way round → $-1 \le x \le 1 \text{ or } x \ge 5$

Answer: $\dfrac{3 x + 3}{2}$

Write $y = \dfrac{2 x - 3}{3}$ and solve for $x\mathclose{:}$ $3y = 2 x - 3\mathclose{,}$ so $x = \dfrac{3 y + 3}{2}\mathclose{.}$ Swap the letters: $f^{-1}(x) = \dfrac{3 x + 3}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the reciprocal $\frac{1}{f(x)}$ instead of the inverse → $\frac{3}{2 x - 3}$
• Undid each step, but in the original order instead of the reverse order → $\frac{3 x}{2} + 9$

Answer: $y = \left(x + 5\right)^{4} + 5$

Shifting 5 units to the left replaces $x$ by $x + 5\mathclose{;}$ shifting 5 units up adds $5\mathclose{:}$ $y = \left(x + 5\right)^{4} + 5\mathclose{.}$

Common mistakes, and the answer each one gives:

• Shifted the wrong way horizontally → $y = \left(x - 5\right)^{4} + 5$
• Shifted the wrong way vertically → $y = \left(x + 5\right)^{4} - 5$

Answer: $9$

Raise both sides to the power $\frac{2}{3}\mathclose{,}$ which undoes $\frac{3}{2}\mathclose{:}$ $x = 27^{\frac{2}{3}} = \left(\sqrt[3]{27}\right)^{2} = 3^{2} = 9\mathclose{.}$

Common mistakes, and the answer each one gives:

• Raised both sides to the power $\frac{3}{2}$ instead of $\frac{2}{3}$ → $81 \sqrt{3}$
• Multiplied by $\frac{2}{3}$ instead of raising to that power → $18$

Answer: $x = 4$

Put $u = \sqrt{x}\mathclose{,}$ so that $x = u^2\mathclose{:}$ $u^2 + u - 6 = (u + 3)(u - 2) = 0\mathclose{,}$ so $u = -3$ or $u = 2\mathclose{.}$ A square root is never negative, so $\sqrt{x} = -3$ is impossible; $\sqrt{x} = 2$ gives $x = 4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Kept the negative value of $\sqrt{x}$ and squared it → $x = 4 \text{ or } x = 9$
• Gave the values of $u = \sqrt{x}$ and stopped → $x = -3 \text{ or } x = 2$

Answer: $-12$

The leading term of a product is the product of the leading terms: $\left(2 x\right)^{2} \cdot (-3x) = 4x^{2} \cdot (-3x) = -12x^{3}\mathclose{.}$ So the leading coefficient is $-12\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not raise the coefficient to its power → $-6$
• Took the leading term of $\left(1 - 3x\right)$ as $+3x$ → $12$

Answer: $x = -3,\ x = -1,\ x = 1 \text{ or } x = 3$

Put $u = x^2\mathclose{:}$ $u^2 - 10u + 9 = (u - 1)(u - 9) = 0\mathclose{,}$ so $x^2 = 1$ or $x^2 = 9\mathclose{.}$ Each gives TWO values of $x\mathclose{:}$ $x = -3,\ x = -1,\ x = 1 \text{ or } x = 3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Solved for $u = x^2$ and stopped → $x = 1 \text{ or } x = 9$
• Kept only the positive square roots → $x = 1 \text{ or } x = 3$

Answer: $y = 2\sqrt{x + 1} + 2$

In that order: the stretch gives $y = 2\sqrt{x}\mathclose{,}$ and the shifts then give $y = 2\sqrt{x + 1} + 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Stretched AFTER shifting up or down, so the vertical shift was stretched too → $y = 2\left(\sqrt{x + 1} + 2\right)$
• Shifted the wrong way horizontally → $y = 2\sqrt{x - 1} + 2$

Answer: $\left(x + 2\right)^{2} - 3 \text{ for } x \ge -2$

Write $y = \sqrt{x + 3} - 2$ and undo the steps in reverse order: take $-2$ off, $y + 2 = \sqrt{x + 3}\mathclose{;}$ square, $\left(y + 2\right)^{2} = x + 3\mathclose{;}$ then $x = \left(y + 2\right)^{2} - 3\mathclose{.}$ Swap the letters: $f^{-1}(x) = \left(x + 2\right)^{2} - 3\mathclose{.}$ A square root is never negative, so $f$ only takes values $y \ge -2$ — and those are the $x$ the inverse is defined for: $x \ge -2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Squared both sides before taking the $-2$ off → $x^{2} - 1$
• Undid the steps in the original order instead of the reverse order → $\left(x - 3\right)^{2} + 2$

Answer: $-3$

A cube is undone by a cube root, and unlike a square root it keeps the sign: $\sqrt[3]{8} = 2\mathclose{.}$ So $x + 5 = 2$ and $x = -3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided by 3 instead of taking the cube root → $- \frac{7}{3}$

Answer: $x = 7$

Square both sides: $x + 18 = \left(x - 2\right)^{2}\mathclose{,}$ so $x^{2} - 5 x - 14 = 0$ and $x = 7$ or $x = -2\mathclose{.}$ SQUARING CAN ADD SOLUTIONS, so check both in the original: at $x = 7$ both sides are $5\mathclose{;}$ at $x = -2$ the right-hand side is $-4\mathclose{,}$ negative, while a square root never is. Only $x = 7$ survives.

Common mistakes, and the answer each one gives:

• Kept both solutions of the squared equation without checking them → $x = -2 \text{ or } x = 7$
• Gave the solution that fails the check → $x = -2$

Answer: $14$

The highest power in the product comes from the highest powers of the factors: $\left(x^{2}\right)^{4} \cdot \left(2x^{3}\right)^{2}$ has degree $8 + 6 = 14\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added only the exponents outside the brackets → $6$
• Multiplied the degrees instead of adding them → $48$

Answer: $x = -5,\ x = 0 \text{ or } x = 1$

Every term has a factor $x\mathclose{,}$ so take it out: $x\left(x^{2} + 4 x - 5\right) = 0\mathclose{.}$ Either $x = 0\mathclose{,}$ or $x^{2} + 4 x - 5 = \left(x + 5\right)\left(x - 1\right) = 0\mathclose{.}$ The solutions are $x = -5,\ x = 0 \text{ or } x = 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided both sides by $x$ and lost the solution $x = 0$ → $x = -5 \text{ or } x = 1$
• Read the solutions off the brackets with the signs unchanged → $x = -1,\ x = 0 \text{ or } x = 5$

Answer: $(2, 4)$

$y = \sqrt{x}$ starts at $(0, 0)\mathclose{,}$ because $x$ may not be negative under the root. The shifts move that starting point to $(2, 4)\mathclose{:}$ the new graph is $y = \sqrt{x - 2} + 4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Shifted the wrong way horizontally → $(-2, 4)$
• Swapped the two coordinates → $(4, 2)$

Answer: $\dfrac{2 x + 8}{4 x + 4}$

Write $y = \dfrac{8 - 4 x}{4 x - 2}$ and solve for $x\mathclose{:}$ $y\left(4 x - 2\right) = 8 - 4 x\mathclose{,}$ so $4 x y - 2 y = 8 - 4 x\mathclose{.}$ Gather the $x$ terms: $x\left(4 y + 4\right) = 2 y + 8\mathclose{,}$ so $x = \dfrac{2 y + 8}{4 y + 4}\mathclose{.}$ Swap the letters: $f^{-1}(x) = \dfrac{2 x + 8}{4 x + 4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the reciprocal $\frac{1}{f(x)}$ instead of the inverse → $\frac{4 x - 2}{8 - 4 x}$
• Swapped $a$ and $d$ but did not change the signs of $b$ and $c$ → $\frac{8 - 2 x}{4 x - 4}$

Answer: $x < -4 \text{ or } -2 < x < 6$

Factorise: $x^{3} - 28 x - 48 = \left(x + 4\right)\left(x + 2\right)\left(x - 6\right)\mathclose{,}$ which is zero at $x = -4\mathclose{,}$ $-2$ and $6\mathclose{.}$ Far to the right the sign is that of the leading term, positive, and it changes at each of the three roots. So the polynomial is positive for $-4 < x < -2 \text{ or } x > 6$ and negative for $x < -4 \text{ or } -2 < x < 6\mathclose{.}$ The answer is $x < -4 \text{ or } -2 < x < 6\mathclose{.}$

Common mistakes, and the answer each one gives:

• Treated it like a quadratic, with only the outer two roots → $-4 < x < 6$
• Took the sign chart the wrong way round → $-4 < x < -2 \text{ or } x > 6$

Answer: $(2, -2)$

$y = \dfrac{1}{x^{2}}$ has the asymptotes $x = 0$ and $y = 0\mathclose{,}$ crossing at the origin. The shifts move every point of the picture, the asymptotes included: the vertical one to $x = 2$ and the horizontal one to $y = -2\mathclose{.}$ They cross at $(2, -2)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Shifted the vertical asymptote the wrong way → $(-2, -2)$
• Kept the asymptotes of $y = \dfrac{1}{x^{2}}$ → $(0, 0)$

Answer: $25$

Raise both sides to the power $\frac{2}{3}\mathclose{,}$ which undoes $\frac{3}{2}\mathclose{:}$ $x = 125^{\frac{2}{3}} = \left(\sqrt[3]{125}\right)^{2} = 5^{2} = 25\mathclose{.}$

Common mistakes, and the answer each one gives:

• Raised both sides to the power $\frac{3}{2}$ instead of $\frac{2}{3}$ → $625 \sqrt{5}$
• Multiplied by $\frac{2}{3}$ instead of raising to that power → $\frac{250}{3}$

Answer: $x = 9 \text{ or } x = 16$

Put $u = \sqrt{x}\mathclose{,}$ so that $x = u^2\mathclose{:}$ $u^2 - 7u + 12 = (u - 3)(u - 4) = 0\mathclose{,}$ so $u = 3$ or $u = 4\mathclose{.}$ Both are positive, so both are possible: $\sqrt{x} = 3$ gives $x = 9$ and $\sqrt{x} = 4$ gives $x = 16\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the values of $u = \sqrt{x}$ and stopped → $x = 3 \text{ or } x = 4$
• Took the square roots of the values of $u$ instead of squaring them → $x = \sqrt{3} \text{ or } x = 2$

Answer: $16$

The leading term of a product is the product of the leading terms: $\left(- 2 x\right)^{3} \cdot (-2x) = -8x^{3} \cdot (-2x) = 16x^{4}\mathclose{.}$ So the leading coefficient is $16\mathclose{.}$

Common mistakes, and the answer each one gives:

• Did not raise the coefficient to its power → $4$
• Took the leading term of $\left(3 - 2x\right)$ as $+2x$ → $-16$

Answer: $x = -5,\ x = -3,\ x = 3 \text{ or } x = 5$

Put $u = x^2\mathclose{:}$ $u^2 - 34u + 225 = (u - 9)(u - 25) = 0\mathclose{,}$ so $x^2 = 9$ or $x^2 = 25\mathclose{.}$ Each gives TWO values of $x\mathclose{:}$ $x = -5,\ x = -3,\ x = 3 \text{ or } x = 5\mathclose{.}$

Common mistakes, and the answer each one gives:

• Solved for $u = x^2$ and stopped → $x = 9 \text{ or } x = 25$
• Kept only the positive square roots → $x = 3 \text{ or } x = 5$

Answer: $\dfrac{2 x - 5}{4}$

Write $y = \dfrac{4 x + 5}{2}$ and solve for $x\mathclose{:}$ $2y = 4 x + 5\mathclose{,}$ so $x = \dfrac{2 y - 5}{4}\mathclose{.}$ Swap the letters: $f^{-1}(x) = \dfrac{2 x - 5}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the reciprocal $\frac{1}{f(x)}$ instead of the inverse → $\frac{2}{4 x + 5}$
• Undid each step, but in the original order instead of the reverse order → $\frac{x}{2} - 10$

Answer: $y = 3\left(x + 4\right)^{3} - 3$

In that order: the stretch gives $y = 3x^{3}\mathclose{,}$ and the shifts then give $y = 3\left(x + 4\right)^{3} - 3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Stretched AFTER shifting up or down, so the vertical shift was stretched too → $y = 3\left(\left(x + 4\right)^{3} - 3\right)$
• Shifted the wrong way horizontally → $y = 3\left(x - 4\right)^{3} - 3$