Functions: domain, range, asymptotes and limits

Answer: $11$

Replace EVERY $x$ by $(-1)\mathclose{:}$ $f(-1) = 4\cdot(-1)^{2} - 3\cdot(-1) + 4 = 4 + 3 + 4 = 11\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took $(-1)^2$ as $-1$ → $3$
• Lost the minus sign of $-1$ in the middle term → $5$

Answer: $(-6, -4) \cup (-4, \infty)$

A logarithm needs a POSITIVE argument: $x + 6 > 0\mathclose{,}$ so $x > -6\mathclose{.}$ The denominator is zero at $x = -4\mathclose{,}$ which is left out: the domain is $(-6, -4) \cup (-4, \infty)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Allowed $x = -6\mathclose{,}$ where $\ln(0)$ is not defined → $[-6, -4) \cup (-4, \infty)$
• Forgot that $x = -4$ makes the denominator zero → $(-6, \infty)$

Answer: $y = 1$

For large $x$ only the leading terms matter: $f(x) \approx \dfrac{2 x}{2 x} = 1\mathclose{.}$ So the horizontal asymptote is $y = 1\mathclose{.}$ (The vertical one is $x = -\dfrac{5}{2}\mathclose{.})$

Common mistakes, and the answer each one gives:

• Gave the vertical asymptote instead → $x = -\dfrac{5}{2}$
• Divided the constant terms instead of the leading coefficients → $y = \dfrac{8}{5}$

Answer: $x = 2$

Factorise both: $f(x) = \dfrac{\left(x - 4\right)\left(x + 6\right)}{\left(x - 4\right)\left(x - 2\right)}\mathclose{.}$ The factor $\left(x - 4\right)$ CANCELS, so $x = 4$ is a perforation, not an asymptote. What is left, $\dfrac{x + 6}{x - 2}\mathclose{,}$ has a denominator that is zero at $x = 2$ while the numerator is not: the vertical asymptote is $x = 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the factor that cancels for an asymptote → $x = 4$
• Set the numerator equal to zero → $x = -6$

Answer: $x^{2} + 2 x + 5$

Divide the leading terms, multiply back and subtract, once per power: $x^{3} \div x = x^{2}\mathclose{;}$ subtract $x^{3} - 4 x^{2}\mathclose{,}$ leaving $2 x^{2} - 3 x + 5\mathclose{.}$ $2 x^{2} \div x = 2 x\mathclose{;}$ subtract $2 x^{2} - 8 x\mathclose{,}$ leaving $5 x + 5\mathclose{.}$ $5 x \div x = 5\mathclose{;}$ subtract $5 x - 20\mathclose{,}$ leaving $25\mathclose{.}$ The quotient is $x^{2} + 2 x + 5$ and the remainder $25\mathclose{:}$ $x^{3} - 2 x^{2} - 3 x + 5 = \left(x - 4\right)\left(x^{2} + 2 x + 5\right) + 25\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided by $x + 4$ instead of $x - 4$ → $x^{2} - 6 x + 21$

Answer: $\dfrac{1}{10}$

At $x = 0$ this is $\frac{0}{0}\mathclose{.}$ Multiply top and bottom by the conjugate $\sqrt{x + 25} + 5\mathclose{:}$ since $\left(\sqrt{x + 25} - 5\right)\left(\sqrt{x + 25} + 5\right) = x\mathclose{,}$ the fraction becomes $\dfrac{x}{x\left(\sqrt{x + 25} + 5\right)} = \dfrac{1}{\sqrt{x + 25} + 5}\mathclose{,}$ which tends to $\dfrac{1}{5 + 5} = \dfrac{1}{10}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Called $\frac{0}{0}$ zero → $0$
• Multiplied by the conjugate but kept only the $5$ of it → $\dfrac{1}{5}$

Answer: $(-8, \infty)$

$5^{x}$ takes every POSITIVE value and never reaches $0\mathclose{.}$ Times $3$ that is still every positive value, and taking $8$ off moves the asymptote to $y = -8\mathclose{,}$ which is approached and never reached: the range is $(-8, \infty)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Included the asymptote itself → $[-8, \infty)$
• Gave the range of $5^{x}$ alone → $(0, \infty)$

Answer: $\dfrac{x + 5}{x - 7}$

Factorise both: $\dfrac{\left(x - 8\right) \left(x + 5\right)}{\left(x - 8\right) \left(x - 7\right)}\mathclose{.}$ The factor $x - 8$ is common to both and cancels, leaving $\dfrac{x + 5}{x - 7}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Cancelled the wrong bracket, keeping the common one → $\frac{x - 8}{x - 7}$
• Cancelled the $x^2$ terms, which are terms and not factors → $\frac{3 x + 40}{15 x - 56}$

Answer: $\dfrac{7}{3}$

Multiply both sides by $x - 2\mathclose{:}$ $3 x - 5 = 6\left(x - 2\right)\mathclose{,}$ which is $3 x - 5 = 6 x - 12\mathclose{.}$ Collect: $- 3 x = -7\mathclose{,}$ so $x = \dfrac{7}{3}\mathclose{.}$ That is not $x = 2\mathclose{,}$ the value the equation forbids, so it stands.

Common mistakes, and the answer each one gives:

• Ignored the denominator and solved $3 x - 5 = 6$ → $\frac{11}{3}$
• Multiplied out $6\left(x - 2\right)$ as $6 x - 2\mathclose{,}$ multiplying only the $x$ → $-1$

Answer: $y = x - 1$

Divide: $x^{2} - 2 x - 4 = \left(x - 1\right)\left(x - 1\right) - 5\mathclose{,}$ so $f(x) = x - 1 - \dfrac{5}{x - 1}\mathclose{.}$ The fraction tends to $0$ as $x$ grows, so the asymptote is $y = x - 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the vertical asymptote instead → $x = 1$
• Divided only the leading terms, $x^2 \div x$ → $y = x$
• Divided by $x$ instead of by $x - 1$ → $y = x - 2$

Answer: $(-\infty, -3)$

The root needs $- x - 3 \ge 0\mathclose{,}$ and because it is in the denominator it may not be zero either: $- x - 3 > 0\mathclose{,}$ so $x < -3\mathclose{:}$ the domain is $(-\infty, -3)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Allowed $x = -3\mathclose{,}$ where the denominator is zero → $(-\infty, -3]$
• Turned the inequality round → $(-3, \infty)$

Answer: $(-\infty, 5)$

A square bracket includes its end and a round one leaves it out; $\infty$ always takes a round one. Here $5$ is left out, and there is no lower end: $(-\infty, 5)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used a square bracket for $<$ → $(-\infty, 5]$
• Pointed the interval the wrong way → $(5, \infty)$

Answer: $[5, 14]$

$f(x) = \left(x - 1\right)^{2} + 5$ has its lowest point at $x = 1\mathclose{,}$ which IS in the domain, so the lowest value is $5\mathclose{.}$ The highest is at an end: $f(-1) = 9$ and $f(4) = 14\mathclose{,}$ so the highest value is $14\mathclose{:}$ the range is $[5, 14]\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used only the values at the ends of the domain → $[9, 14]$
• Gave the range on the whole real line → $[5, \infty)$

Answer: $[4, 7) \cup (7, \infty)$

The root needs $x - 4 \ge 0\mathclose{,}$ so $x \ge 4\mathclose{,}$ and the denominator is zero at $x = 7\mathclose{,}$ which is left out: the domain is $[4, 7) \cup (7, \infty)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot that $x = 7$ makes the denominator zero → $[4, \infty)$
• Required the inside of the root to be strictly positive → $(4, 7) \cup (7, \infty)$

Answer: $x = -1$

The vertical asymptote sits where the denominator is zero: $x + 1 = 0\mathclose{,}$ so $x = -1\mathclose{.}$ (The horizontal one is $y = -2\mathclose{,}$ the ratio of the leading coefficients.)

Common mistakes, and the answer each one gives:

• Set the NUMERATOR to zero (that gives the $x$-intercept) → $x = 2$
• Gave the horizontal asymptote instead → $y = -2$
• Lost the minus sign when solving the denominator → $x = 1$

Answer: $(-1, 1)$

Factorise the numerator: $x^{2} + 3 x + 2 = \left(x + 1\right)\left(x + 2\right)\mathclose{.}$ The factor $\left(x + 1\right)$ cancels, so $f(x) = x + 2$ everywhere EXCEPT at $x = -1\mathclose{,}$ where the original is $\frac{0}{0}$ and undefined. The hole is at $x = -1\mathclose{,}$ at the height the simplified formula gives there: $y = -1 + 2 = 1\mathclose{.}$ So it is $(-1, 1)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Read the hole off the factor $\left(x + 1\right)$ with the wrong sign → $(1, 3)$
• Gave the point where the graph meets the $x$-axis → $(-2, 0)$

Answer: $-9$

$x^{3} \div x = x^{2}\mathclose{;}$ subtract $x^{3} + x^{2}\mathclose{,}$ leaving $- 7 x^{2} + x - 1\mathclose{.}$ $- 7 x^{2} \div x = - 7 x\mathclose{;}$ subtract $- 7 x^{2} - 7 x\mathclose{,}$ leaving $8 x - 1\mathclose{.}$ $8 x \div x = 8\mathclose{;}$ subtract $8 x + 8\mathclose{,}$ leaving $-9\mathclose{:}$ the quotient is $x^{2} - 7 x + 8$ and the remainder $-9\mathclose{.}$ Check it with the remainder theorem: the remainder on dividing by $x + 1$ is the value at $x = -1\mathclose{,}$ and $f(-1) = -9\mathclose{.}$

Common mistakes, and the answer each one gives:

• Put in $x = 1$ instead of $x = -1$ → $-5$
• Gave the constant term of the quotient → $8$

Answer: $-\dfrac{1}{3}$

Divide every term by $x^2\mathclose{,}$ the highest power: the fraction is $\dfrac{-2 - \frac{9}{x} - \frac{8}{x^2}}{6 - \frac{3}{x} + \frac{6}{x^2}}\mathclose{,}$ and every term with $x$ underneath tends to $0\mathclose{.}$ The limit is $\dfrac{-2}{6} = -\dfrac{1}{3}\mathclose{,}$ the graph's horizontal asymptote.

Common mistakes, and the answer each one gives:

• Divided the constant terms → $-\dfrac{4}{3}$
• Divided the coefficients of $x$ → $3$

Answer: $[4, \infty)$

Complete the square: $f(x) = 2\left(x - 3\right)^{2} + 4\mathclose{.}$ The vertex is $(3, 4)$ and the parabola opens upwards, so its lowest value is $4\mathclose{,}$ and every value above it is taken: the range is $[4, \infty)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the $x$-coordinate of the vertex as the bound → $[3, \infty)$
• Took the parabola to open the other way → $(-\infty, 4]$

Answer: $(-\infty, -6) \cup (-6, 1]$

The root needs $1 - x \ge 0\mathclose{,}$ so $x \le 1\mathclose{.}$ The denominator is zero at $x = \pm 6\mathclose{,}$ and those values are left out: the domain is $(-\infty, -6) \cup (-6, 1]\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot the values that make the denominator zero → $(-\infty, 1]$
• Turned the root's condition round, taking $x \ge 1$ → $[1, 6) \cup (6, \infty)$

Answer: $\dfrac{x - 7}{x - 4}$

Factorise both: $\dfrac{\left(x - 7\right) \left(x + 5\right)}{\left(x - 4\right) \left(x + 5\right)}\mathclose{.}$ The factor $x + 5$ is common to both and cancels, leaving $\dfrac{x - 7}{x - 4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Cancelled the wrong bracket, keeping the common one → $\frac{x + 5}{x - 4}$
• Cancelled the $x^2$ terms, which are terms and not factors → $\frac{- 2 x - 35}{x - 20}$

Answer: $-\dfrac{25}{3}$

Multiply both sides by $x + 6\mathclose{:}$ $2 x + 5 = 5\left(x + 6\right)\mathclose{,}$ which is $2 x + 5 = 5 x + 30\mathclose{.}$ Collect: $- 3 x = 25\mathclose{,}$ so $x = -\dfrac{25}{3}\mathclose{.}$ That is not $x = -6\mathclose{,}$ the value the equation forbids, so it stands.

Common mistakes, and the answer each one gives:

• Ignored the denominator and solved $2 x + 5 = 5$ → $0$
• Multiplied out $5\left(x + 6\right)$ as $5 x + 6\mathclose{,}$ multiplying only the $x$ → $- \frac{1}{3}$

Answer: $y = x - 2$

Divide: $x^{2} - 9 = \left(x + 2\right)\left(x - 2\right) - 5\mathclose{,}$ so $f(x) = x - 2 - \dfrac{5}{x + 2}\mathclose{.}$ The fraction tends to $0$ as $x$ grows, so the asymptote is $y = x - 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the vertical asymptote instead → $x = -2$
• Divided only the leading terms, $x^2 \div x$ → $y = x$

Answer: $[-5, \infty)$

A square root is never negative, and $\sqrt{x - 3}$ takes every value from $0$ upwards (from $x = 3$ on). Times $3$ that is still every value from $0$ up, and taking $5$ off moves the lowest value to $-5\mathclose{:}$ the range is $[-5, \infty)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the domain instead of the range → $[3, \infty)$
• Forgot that a square root is never negative → $(-\infty, \infty)$

Answer: $y = 3$

For large $x$ only the leading terms matter: $f(x) \approx \dfrac{3 x}{x} = 3\mathclose{.}$ So the horizontal asymptote is $y = 3\mathclose{.}$ (The vertical one is $x = -7\mathclose{.})$

Common mistakes, and the answer each one gives:

• Gave the vertical asymptote instead → $x = -7$
• Divided the constant terms instead of the leading coefficients → $y = \dfrac{6}{7}$

Answer: $x = 7$

Factorise both: $f(x) = \dfrac{\left(x - 3\right)\left(x + 7\right)}{\left(x - 3\right)\left(x - 7\right)}\mathclose{.}$ The factor $\left(x - 3\right)$ CANCELS, so $x = 3$ is a perforation, not an asymptote. What is left, $\dfrac{x + 7}{x - 7}\mathclose{,}$ has a denominator that is zero at $x = 7$ while the numerator is not: the vertical asymptote is $x = 7\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the factor that cancels for an asymptote → $x = 3$
• Set the numerator equal to zero → $x = -7$

Answer: $x^{2} - 3 x + 7$

Divide the leading terms, multiply back and subtract, once per power: $x^{3} \div x = x^{2}\mathclose{;}$ subtract $x^{3} + 3 x^{2}\mathclose{,}$ leaving $- 3 x^{2} - 2 x\mathclose{.}$ $- 3 x^{2} \div x = - 3 x\mathclose{;}$ subtract $- 3 x^{2} - 9 x\mathclose{,}$ leaving $7 x\mathclose{.}$ $7 x \div x = 7\mathclose{;}$ subtract $7 x + 21\mathclose{,}$ leaving $-21\mathclose{.}$ The quotient is $x^{2} - 3 x + 7$ and the remainder $-21\mathclose{:}$ $x^{3} - 2 x = \left(x + 3\right)\left(x^{2} - 3 x + 7\right) - 21\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided by $x - 3$ instead of $x + 3$ → $x^{2} + 3 x + 7$

Answer: $1$

Putting $x = 3$ straight in gives $\frac{0}{0}\mathclose{,}$ which is not a number, so simplify first: $x^{2} - 5 x + 6 = \left(x - 3\right)\left(x - 2\right)\mathclose{,}$ and for $x \ne 3$ the fraction is $x - 2\mathclose{.}$ As $x \to 3$ that tends to $3 - 2 = 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Called $\frac{0}{0}$ zero → $0$
• Simplified correctly, then put in $x = -3$ → $-5$

Answer: $(-3, 6) \cup (6, \infty)$

A logarithm needs a POSITIVE argument: $x + 3 > 0\mathclose{,}$ so $x > -3\mathclose{.}$ The denominator is zero at $x = 6\mathclose{,}$ which is left out: the domain is $(-3, 6) \cup (6, \infty)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Allowed $x = -3\mathclose{,}$ where $\ln(0)$ is not defined → $[-3, 6) \cup (6, \infty)$
• Forgot that $x = 6$ makes the denominator zero → $(-3, \infty)$

Answer: $-4$

Replace EVERY $x$ by $(-1)\mathclose{:}$ $f(-1) = 3\cdot(-1)^{2} + 2\cdot(-1) - 5 = 3 - 2 - 5 = -4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took $(-1)^2$ as $-1$ → $-10$
• Lost the minus sign of $-1$ in the middle term → $0$