Introduction

No matter how well you understand this course, mistakes happen. Some may come from rushing, others from mixing up notation. What’s frustrating is that the majority of points off will always come from small mistakes that completely change the result. Here, we will focus on some of the most seen mistakes throughout the course. The goal by the end of this is to not just know what to do and what not to, but to understand why each mistake is wrong. Without further ado, let’s begin!

The Freshman’s Dream

This mistake is so infamous that it received its own name: assuming that ${\left(a+b\right)}^2=a^2+b^2$.

Squaring a binomial means multiplying the entire expression by itself:

${\left(a+b\right)}^2=\left(a+b\right)\left(a+b\right)$

Using the distributive property:

$\left(a+b\right)\left(a+b\right)=a^2+ab+ba+b^2=a^2+2ab+b^2$

The missing middle term is $2ab$. For example:

${\left(x+5\right)}^2=x^2+10x+25$, not $x^2+25$

If you want to get a refresher on this topic, we have a lesson on the Binomial Theorem.

Confusing Inverse Trig Notation

This happens when ${\sin }^{-1}\left(x\right),{\cos }^{-1}\left(x\right),$ and ${\tan }^{-1}\left(x\right)$ are treated as $\frac{1}{\sin \left(x\right)},\frac{1}{\cos \left(x\right)},$ or $\frac{1}{\tan \left(x\right)}$.

In algebra, a $-1$ exponent usually means a reciprocal $\left(x^{-1}=\frac{1}{x}\right)$.

However, in trigonometry, the $-1$ superscript is standard notation for the inverse function (also written as “arc”):

${\sin }^{-1}\left(x\right)=\arcsin \left(x\right)\ne \frac{1}{\sin \left(x\right)}$

The reciprocal of $\sin \left(x\right)$ is the cosecant function:

$\csc \left(x\right)=\frac{1}{\sin \left(x\right)}={\left(\sin \left(x\right)\right)}^{-1}$

Notice how there is an extra set of parentheses around the $\sin$ function. This is how we know to take the reciprocal instead of the inverse.

For more info about inverse trig functions, please look at this article.

Distributing the Negative

An error in distributing the negative occurs when parentheses are dropped or only the first term is negated when subtracting polynomials.

When subtracting an entire expression, such as $f\left(x\right)-g\left(x\right)$ where $g\left(x\right)=2x^2-3x+4$:

Incorrect: $-\left(2x^2-3x+4\right)=-2x^2-3x+4$

Correct: $-\left(2x^2-3x+4\right)=-2x^2+3x-4$

Domain Restrictions

Remember that certain functions or simplifications have certain rules on their domains.

Consider the function $f\left(x\right)=\frac{x^2-4}{x-2}$.

Factoring gives us $f\left(x\right)=\frac{\left(x-2\right)\left(x+2\right)}{x-2}=x+2\text{ for }x\ne 2$.

Even though the factor $\left(x-2\right)$ cancels, the original function is undefined at $x=2$. It produces a removable discontinuity (hole) at $\left(2,4\right)$, not a continuous line.

Some Domain Rules to Remember:

Radians and Degrees

When you walk into the room for your AP Exam, one of the first things you should do is check to make sure your calculator is in radians.

In AP Precalculus, all trigonometry assumes radians unless explicitly stated otherwise with a degree symbol $\left(°\right)$.

Radians and Degree Conversion

Be careful to multiply by the correct conversion factor:

Degrees to Radians: Multiply by $\frac{\pi }{180°}$

$150°\times \frac{\pi }{180°}=\frac{5\pi }{6}$

Radians to Degrees: Multiply by $\frac{180°}{\pi }$

$\frac{3\pi }{4}\times \frac{180°}{\pi }=135°$

Exponent and Logarithm Properties

Make sure to use the correct properties of powers or logs.

$\begin{array}{cc}\text{Incorrect}& \text{Correct}\\ \ln \left(a+b\right)=\ln \left(a\right)+\ln \left(b\right)& \ln \left(a\cdot b\right)=\ln \left(a\right)+\ln \left(b\right)\\ \frac{\ln \left(a\right)}{\ln \left(b\right)}=\ln \left(a-b\right)& \ln \left(\frac{a}{b}\right)=\ln \left(a\right)-\ln \left(b\right)\\ {\left(\ln \left(x\right)\right)}^2=2\ln \left(x\right)& \ln \left(x^2\right)=2\ln \left(x\right)\left(\text{for }x>0\right)\\ e^{a+b}=e^a+e^b& e^{a+b}=e^a{e}^b\end{array}$

Cancelling

You can only cancel factors that multiply the entire numerator and denominator:

$\frac{x+6}{x}\ne 1+6=7$

The denominator must divide each term:

$\frac{x+6}{x}=\frac{x}{x}+\frac{6}{x}=1+\frac{6}{x}$

Square Roots

Square roots do not distribute across addition, meaning that $\sqrt{x^2+9}\ne x+3$.

To confirm this, take a look at $x=4$:

$\sqrt{4^2+9}=\sqrt{16+9}=\sqrt{25}=5\ne 4+3=7$

Additionally, you must remember to include $\pm$ when taking square roots to solve an equation to account for additional solutions:

$x^2=16\to x=\pm 4$

Note: The radical symbol $\sqrt{16}$ alone refers strictly to the principal root, $+4$. The $\pm$ is required when taking the square root of both sides of an equation.

Rationalizing the Denominator

In algebra, we tend to dislike having radicals in the denominator. Omitting rationalization when simplest form is requested may result in points off.

Single Radical Term:

$\frac{5}{\sqrt{3}}=\frac{5}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{5\sqrt{3}}{3}$

Binomial Denominator (Use the Conjugate):

$\frac{2}{3-\sqrt{5}}=\frac{2}{3-\sqrt{5}}\cdot \frac{3+\sqrt{5}}{3+\sqrt{5}}=\frac{2\left(3+\sqrt{5}\right)}{3^2-{\left(\sqrt{5}\right)}^2}=\frac{6+2\sqrt{5}}{9-5}=\frac{6+2\sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}$

Slant and Oblique Asymptotes

Slant and Oblique asymptotes are not the same.

An Oblique Asymptote is any non-horizontal polynomial asymptote describing end behavior as $x\to \pm \infty$ when the degree of the numerator is strictly greater than the degree of the denominator.

A Slant Asymptote is a specific type of oblique asymptote that is a linear function of the form $y=mx+b\left(m\ne 0\right)$. This occurs only when the numerator degree is exactly one greater than the denominator degree.

If $f\left(x\right)=\frac{x^3+2x-1}{x+1}=x^2-x+3-\frac{4}{x+1}$, the end behavior is modeled by the parabolic asymptote $y=x^2-x+3$. It is an oblique (nonlinear) asymptote, but it is not a slant asymptote.

Semi-Log Graphs

Be careful to not misinterpret semi-logarithmic plots (where the vertical axis is scaled by powers of $10$ or $e$).

An exponential function $y=a\cdot b^x$ appears as a straight line on a semi-log plot (where $\log \left(y\right)$ is plotted against $x$).

$\log \left(y\right)=\log \left(a\right)+x\log \left(b\right)$

Moving up one tick mark on a log scale represents multiplying by a constant factor (e.g., the first mark is $10^1$, the second is $10^2$, the third is $10^3$), not adding a constant amount.

Additionally, note the difference between a logarithmically scaled $x$-axis and a logarithmically scaled $y$-axis. A logarithm scaled $x$-axis will cause a logarithmic function to appear linear, and a logarithmically scaled $y$-axis will cause exponential functions to appear linear.

Periodicity

Do not provide a single angle when asked for all solutions to a trigonometric equation.

Trigonometric functions are periodic. Unless the problem restricts the domain (e.g., $0\le x<2\pi$), you must write the general solution:

$\sin \left(x\right)=\frac{1}{2}$

$x=\frac{\pi }{6}+2\pi k$ or $x=\frac{5\pi }{6}+2\pi k,k\in \mathbb{Z}$

Furthermore, take note that reciprocal functions’ periods are identical.

Rounding

Unless instructed otherwise, always round to $3$ decimal places.

Intermediate Values: Store numbers in calculator variables ($A,B,C$, etc.). Do not round intermediate steps.

Both rounding and truncating to three decimal places are accepted (e.g., $3.14159\cdots \to 3.142\text{or }3.141$).

Extraneous Solutions

When solving an equation, you may produce a solution outside of the domain of the original equation. This happens most in Logarithmic and Rational Equations.

${\log }_b\left(u\right)$ requires $u>0$.

${\log }_2\left(x\right)+{\log }_2\left(x-2\right)=3\to {\log }_2\left(x\left(x-2\right)\right)=3\to x^2-2x=8$

$\left(x-4\right)\left(x+2\right)=0\to x=4\text{or }x=-2$

Plugging $x=-2$ back into ${\log }_2\left(x\right)$ is undefined, making it extraneous.

Be sure to show that you checked for extraneous solutions!

For Rational Equations, solutions that make the denominator equal to zero are invalid.