Mistakes regarding algebra

A lot of mistakes that people make in Calculus originate in simple algebraic mistakes. Most textbooks of calculus have a section in the beginning dedicated solely to the properties of real numbers. One is not required go through each of the proofs. However it's important to read them at least once because mistakes made when operating with functions often arise from mistakes when operating with the real numbers themselves.

Regarding intuition

• [math]\displaystyle{ \sqrt{1/10} }[/math] vs. [math]\displaystyle{ (1/10)^2 }[/math]. When we calculate [math]\displaystyle{ x^2 }[/math] the number always become larger, except for [math]\displaystyle{ 0 \lt x \lt 1 }[/math]. When we calculate [math]\displaystyle{ \sqrt{x} }[/math] the number always become smaller, except for [math]\displaystyle{ 0 \lt x \lt 1 }[/math]. This causes some problems when evaluating limits and/or plotting the graph of functions because we can be mislead and think that the function is increasing or decreasing when it's not.

Regarding identities

• [math]\displaystyle{ \sqrt{x \pm y} \neq \sqrt{x} \pm \sqrt{y} }[/math]. The root of the sum | difference is not equal to the sum | difference of the roots

• [math]\displaystyle{ \sqrt{x^2 \pm y^2} \neq \sqrt{(x \pm y)^2} }[/math]. Sometimes people cancel out the squares with the roots.

• [math]\displaystyle{ (x + y)^2 = x^2 + 2xy + y^2 }[/math].

• [math]\displaystyle{ (x - y)^2 = x^2 - 2xy + y^2 }[/math].

• [math]\displaystyle{ x^2 - y^2 = (x + y)(x - y) }[/math]. This identity shows up a lot when calculating limits. Do not confuse it with the above two!

• [math]\displaystyle{ e^{{2x}^2 \neq (e^{2x})^2} }[/math]. When exercises have composite functions with powers this mistake is common. Be careful!

• [math]\displaystyle{ \sqrt{x^2} = |x| }[/math]. Be careful with this property. In many cases we want to avoid negative values and one of such ways is to take the absolute value. However, when dealing with differentiation the absolute value brings its own issue because it creates a point at which the derivative doesn't exist. For example: in numerical methods there are problems which require derivatives to calculate distances. A negative distance doesn't exist and there are cases in which we want to square the expression to avoid negative distances.

• [math]\displaystyle{ -x^2 \neq x^2 }[/math] but [math]\displaystyle{ -x^3 = (-x)^3 }[/math]. This quirk is a mistake that happens just about everywhere.

Regarding inequalities

• [math]\displaystyle{ 2 \gt 1 }[/math]. Now multiply by -1 and we have [math]\displaystyle{ -2 \lt -1 }[/math]. It's common for people to forget this fact and make mistakes in calculus because of this. When we have numbers it's easy to quickly see that [math]\displaystyle{ -2 \gt -1 }[/math] is wrong. However, when we have functions on both sides we can easily be fooled and make this mistake.

• [math]\displaystyle{ \frac{3x \ - \ 1}{x \ + \ 2} \geq 5 }[/math]. The inequality reads "For which values of [math]\displaystyle{ x }[/math] we have a function whose graph is above or equal to 5?". First [math]\displaystyle{ x + 2 }[/math] cannot be zero, else we have a division by zero. Now be careful with the comparison! When we have an equality we can multiply both sides and keep the equality the same. That's not the case with inequalities.
  

Let's multiply both sides by [math]\displaystyle{ (x + 2) }[/math] and solve for [math]\displaystyle{ x }[/math]. [math]\displaystyle{ 3x - 1 \geq 5x + 10 }[/math]. We should get [math]\displaystyle{ x \geq -\frac{11}{2} }[/math]

Our analysis is not over yet! We have a quotient of functions. What happens if both the numerator and denominator are negative? That means that for a certain range of negative values we also have that the result is greater than or equal to 5. So [math]\displaystyle{ (x + 2) }[/math] is negative for any [math]\displaystyle{ x \lt -2 }[/math]. [math]\displaystyle{ (3x -1) }[/math] is negative for any [math]\displaystyle{ x \lt 1/3 }[/math]. Conclusion, both are negative when [math]\displaystyle{ x \lt -2 }[/math].

Now combine both results in this form [math]\displaystyle{ \{x \in \mathbb{R} | -\frac{11}{2} \leq x \lt -2\} }[/math] or x is in [math]\displaystyle{ [-5.5, -2[ }[/math]