Mistakes regarding functions

Concerning the concept

• It's easy to confuse the domain of composite functions. As a general rule, the domain of [math]\displaystyle{ f(g(x)) }[/math] can never be larger than the domains of [math]\displaystyle{ f }[/math] or [math]\displaystyle{ g }[/math]. Think about this analogy: the set of natural numbers is a subset of the real numbers. It's impossible for a subset to contain more elements than the larger set which the former is a subset of. Always think on which values are allowed and which ones aren't for the outermost function.

Concerning confusion between single variable and multivariable functions

• I have no idea if this mistake happens, but it's a possibility nonetheless. Suppose that we have [math]\displaystyle{ f(x,y) = x + y }[/math]. It's possible that one sees [math]\displaystyle{ g(x) = x }[/math] and [math]\displaystyle{ h(y) = y }[/math], then [math]\displaystyle{ f(x,y) = g(x) + h(y) }[/math]. The numerical results are the same for both the two variable function and the sum of two single variable functions. However, we can't just split a multivariable function like that! By doing so we are, inadvertently, considering that we have different functions governing different processes. Each function having its own particular domain. One practical case is to consider the square root or trigonometric functions. For some pairs of numbers it may hold true, but we can quickly see that a sum of roots or sum of sines won't be the same as the sine of the sum or the root of the sum.

• [math]\displaystyle{ f(x,y) = x^2 }[/math] is this a single variable function or a multivariable one? It's multivariable. The fact that the equation doesn't have [math]\displaystyle{ y }[/math] is not the same as to say that "the second variable doesn't exist". It's there. Where? It's a constant, the null (zero) constant.

• [math]\displaystyle{ f(a,b) = ab }[/math] is a function of two variables. However, [math]\displaystyle{ f(x) = (a + b)x }[/math] is a function of one variable. Sometimes this quirk happens and we see more variables than we should because we make such a strong association between letters being variables that we forget that sometimes they are constants. This is specially common when we have expressions with multiple parameters or variables in physics but only one is important when we are differentiating or integrating.

Concerning properties

• [math]\displaystyle{ f(x) = 3 }[/math] is a constant function and the number 3 is odd. But this function is even. It's a bit misleading.

Concerning the zero and the empty set

• I don't know if this confusion happens, but people may think that if a function is undefined at a point, it's the same as to say that the function is equal to zero there. Not quite the same thing. Zero is a number and if a set contains the zero alone, it's not an empty set because it contains something.

Concerning trigonometry

• A mistake that can happen is for people to think that when we have a function that is a combination of trig and non-trig functions, the arguments are different. No!! Remember, a radian is just some special case of irrational numbers. There is no need to plug something different in the square root, log or polynomial. I think that a related confusion is because sometimes we read [math]\displaystyle{ \pi }[/math] as [math]\displaystyle{ 1 \pi }[/math], which can cause the confusion between [math]\displaystyle{ \pi }[/math] and one radian.

Concerning signs

• Let [math]\displaystyle{ x_1 }[/math] [math]\displaystyle{ x_2 }[/math] be two numbers from the function's domain such that [math]\displaystyle{ x_2 \gt x_1 }[/math]. We cannot conclude that [math]\displaystyle{ f(x_2) \gt f(x_1) }[/math] without knowing whether the function is crescent or not! A similar confusion happens when we think that [math]\displaystyle{ -2 \gt -1 }[/math] because the negative sign inverts the comparison.

• Suppose we have this function [math]\displaystyle{ f(x) = -x }[/math]. If we input [math]\displaystyle{ -x }[/math], the function outputs a positive value, not a negative one! For example: we wish to calculate [math]\displaystyle{ f(-2) }[/math]. Due to some quirk in our minds we confuse the dependent and independent variables and blindly do this [math]\displaystyle{ f(-2) = -2 }[/math]. My hypothesis is that, mentally, we often look at [math]\displaystyle{ -x }[/math] and [math]\displaystyle{ (-1)x }[/math] and think that they are different from each other.

• [math]\displaystyle{ f^{-1}(x) \neq \frac{1}{f(x)} }[/math]. The meaning of [math]\displaystyle{ f^{-1} }[/math] is the inverse of the function, which is not the same as calculating the inverse of the value of the function at [math]\displaystyle{ x }[/math]. It's often confusing, but the inverse of a function is a concept. The inverse of a number is another number, usually we think on a fraction. Now a function itself is not a number and the inverse function is also not a number, but the concept that if we can relate [math]\displaystyle{ a }[/math] to [math]\displaystyle{ b }[/math], we can make the reversed relationship, [math]\displaystyle{ b }[/math] to [math]\displaystyle{ a }[/math]. That's how I'd try to differentiate using the same word and notation to mean two different things.
  
  [math]\displaystyle{ -f(x) }[/math] on the other hand means to multiply the function by minus one. This operation essentially mirrors all values of the function at each point. Positive becomes negative. Negative becomes positive. Careful! We are not inverting the sign of the function's argument!

• There is another related confusion that is caused by the words "inverse" and "opposite". Ask somebody what is the inverse / opposite of going up? The person is going to answer that is to go down. In daily life we can interchange both words without causing any harm. In mathematics, however, there is a problem. The inverse of exp is log, but both are functions that are always increasing or decreasing if you invert the sign to negative. There is the confusion! It's not uncommon for people to think that the inverse of a function is a function that is, literally, decreasing if the other is increasing and vice-versa.

Concerning equations

• We often learn that [math]\displaystyle{ x^2 - 1 = 0 }[/math] and [math]\displaystyle{ 2x^2 - 2 = 0 }[/math] have the same roots. When solving equations we can multiply by any constant factor because it doesn't change the roots. However, when plotting the function we cannot do that! To multiply a function by a constant factor changes its graph, hence it's no longer the same function. Think about vectors and linear algebra. When we multiply a vector by a constant, we keep the same orientation and direction but change its magnitude. For example: take the function [math]\displaystyle{ \sin(x) }[/math] or [math]\displaystyle{ \cos(x) }[/math], if we multiply it by a constant, we are changing the function's amplitude but not the points where the graph crosses the x axis.

• When one is learning functions for the first time, the expression [math]\displaystyle{ 1 = 2 }[/math] is never true. However, [math]\displaystyle{ f(1) = 2 }[/math] can be true and that depends on the function. It reads as "function [math]\displaystyle{ f }[/math], calculated at the point [math]\displaystyle{ x = 1 }[/math], is equal to 2" or "the function [math]\displaystyle{ f }[/math] is equal to 2 when we calculate it for [math]\displaystyle{ x = 1 }[/math]".

• Going further in calculus one can find exercises with expressions similar to this: [math]\displaystyle{ f(x^2 + 1) = g(x) }[/math]. Let's call [math]\displaystyle{ h(x) = x^2 + 1 }[/math] then what we have is this: [math]\displaystyle{ f(h(x)) = g(x) }[/math]. Be careful to not confuse it with [math]\displaystyle{ x^2 + 1 = x }[/math] in the same way as above with [math]\displaystyle{ 1 = 2 \ ! }[/math] The equation has functions on both sides. We aren't looking at an equation were the arguments of each function are equal to each other. What the equality is doing is comparing functions.

• Suppose we have [math]\displaystyle{ f(x) = x^2 }[/math] and [math]\displaystyle{ g(x) = x^3 }[/math]. The sum is [math]\displaystyle{ f(x) + g(x) = x^2 + x^3 }[/math]. It can happen that people confuse the argument with the equation like this: [math]\displaystyle{ f(x) + g(x) = (f + g)(x + x) }[/math]. Who said that to sum functions is the same as to add up their respective arguments? A linear transformation looks similar, except that both sides of the equation have the same function! Not two different functions as in here.

• There is a certain type of equation that we don't learn in calculus and it's a functional equation. It's an equation where the unknown is a function itself. Such as [math]\displaystyle{ f(x^2 + 2) = x^4 + 4 }[/math]. If we know that, can we find [math]\displaystyle{ f(x) = \ ? }[/math].

Concerning argument, dependent x independent variables

• Careful with the confusion between a function's argument and the value of the function itself! Let's say we have [math]\displaystyle{ f(x) = x^2 }[/math]. Now what is the difference between [math]\displaystyle{ f(x + 1) = x^2 }[/math] and [math]\displaystyle{ f(x) = (x + 1)^2 \ ? }[/math] In the first case we have a composite function, we have some [math]\displaystyle{ g(x) = x + 1 }[/math] and for every [math]\displaystyle{ x }[/math] that we are going to calculate, first we calculate [math]\displaystyle{ g(x) }[/math], then use the value that [math]\displaystyle{ g(x) }[/math] outputs as the input for [math]\displaystyle{ f(x) }[/math]. We have [math]\displaystyle{ f(g(x)) = x^2 }[/math].
  
  Now for the second case what we have is another function that is not equal to the first. It's pretty obvious that [math]\displaystyle{ x^2 \neq (x + 1)^2 }[/math]. Ergo, these are two different functions. When we are careless and clueless we do this [math]\displaystyle{ f(x + 1) = (x + 1)^2 }[/math]. Who said that [math]\displaystyle{ x = x + 1 \ ?? }[/math] In the previous example, we defined [math]\displaystyle{ g(x) = x + 1 }[/math] and not [math]\displaystyle{ g(x) = x }[/math].
  
  Can we have [math]\displaystyle{ f(x) = g(x)^2 \ ? }[/math] Sure, why not? We can define a function as being the square of a different function. Now, is that a composite function? No. We aren't using one function as the argument of another. What we are doing is saying that the value of the function [math]\displaystyle{ f }[/math], at every point, is going to be the value of the function [math]\displaystyle{ g }[/math], squared.

• This [math]\displaystyle{ \overrightarrow{r}(t) = \lt x(t), \ y(t), \ z(t) \gt }[/math] is not a function of three variables. It's a vector valued function. In this case, it associates time with position in space. In physics, motion in 3D space is not described by just one function. It's one different function for each axis, but one independent variable for all.

• [math]\displaystyle{ \frac{\sin(2x)}{2} \neq \sin \left(\frac{1}{2}2x\right) }[/math]. Sometimes people are tempted to cancel out the two, but we can't just do that! A similar mistake is [math]\displaystyle{ 4 \left( \frac{1}{2} \right)^x \neq 2(1)^x }[/math]. I believe that this is caused by the fact that some functions allow us to do that because of this property [math]\displaystyle{ cf(x) = f(cx) }[/math].

• [math]\displaystyle{ f(a,b, ..., z) }[/math]. In linear algebra and analytical geometry we learn that the coordinate's order does matter. It my happen that some people accidentaly swap out the order of them. For ex: [math]\displaystyle{ f(2,3) \neq f(3,2) }[/math]. Relating to that, when we have piecewise functions, there is nothing forbidding the function from defining [math]\displaystyle{ ... \text{se} \ (x,y) = (x,0) }[/math], which means "every ordered pair where [math]\displaystyle{ x }[/math] is anything and [math]\displaystyle{ y = 0 }[/math]".