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The variable(s) in an algebraic function can appear in the numerator, denominator, or under a root symbol, but the exponents of all variables in the function must be rational numbers (fractions or integers).
A problem involving algebraic functions might look something like this:
You know that the age of your uncle John is twice your age plus four years, and you know that your age is 15. You can use an algebraic function to work out the age of your uncle (x): \(x = 15 \cdot 2 +4 = 34\)
In this article, we will define what algebraic functions are, the different types of algebraic functions, how to identify algebraic and non-algebraic functions, touch on the differential calculus of algebraic functions, and work through some examples.
- What are algebraic functions?
- Types of algebraic functions
- Algebraic functions vs. non-algebraic functions
- Differential calculus with algebraic functions
- Graphing algebraic functions
- Algebraic functions example problems
What are Algebraic Functions?
As we learned from our Functions article, there are many classes of functions. One of those classes is algebraic functions.
An Algebraic Function is a function that involves only algebraic operations: addition, subtraction, multiplication, division, powers, and roots.
This class of functions were generated when quotients and fractional powers were included apart from polynomial functions. If it weren't for these allowances, we would simply have polynomial functions! These additions to the polynomial functions give rise to the types of algebraic functions.
Types of algebraic functions
- Polynomial Functions: A polynomial function is a function that is a sum of terms, each of which is (a constant x a nonnegative integer power of the variable). Polynomial functions can be expressed in the form \(f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_2x^2 + a_1x + a_0\), where an, an-1, ..., a2, a1, a0are constants, and n is a nonnegative integer. An example of a polynomial function is \(f(x) = 3x^3 - 2x^2 + x + 5\).
Note that the constant can be a negative number (e.g. -2) so that instead of a sum, the term appears to be a substraction. If the power of the variable were negative, it would mean that the term is in a fraction, and thus would be a rational function and not a polynomial function. When a variable term (x) appears without a constant, it means the constant is actually the number 1 (\(x \cdot 1 = x\)). If a constant appears alone, it means the power of the variable is 0 (\(a \cdot x^0 = a \cdot 1 = a\)).
- Rational Functions: A rational function is a ratio of two polynomial functions. It can be expressed in the form \(f(x) = \frac{p(x)}{q(x)}\), where p(x) and q(x) are polynomial functions. The function is undefined when q(x) = 0, since you cannot divide by zero. An example of a rational function is \(f(x) = \frac{(x^2 - 1)}{(x + 2)} = (x^2-1)(x+2)^{-1}\).
- Power/Root Functions: A power function is a function of the form \(f(x) = kx^n\), where k and n are constants. When n is a fraction, we get a root function. For example, if \(n = \frac{1}{2}\), we get the square root function; if \(n = \frac{1}{3}\), we get the cube root function, and so on. An example of a power function is \(f(x) = 5x^3\). An example of a root function is \(f(x) = \sqrt x\) or \(f(x) = \sqrt[3]x\).
An equation in the form \(n^x\) is NOT an algebraic function, as the variable (x) is the exponent instead of the base.
Examples of Algebraic Functions
Based on our definition of an algebraic function, let's list some examples of algebraic functions.
\[f(x)=3x+2\]
\[g(x)=x^2-2x-9\]
\[h(x)=\sqrt[3]{x}\]
\[j(x)=\dfrac{x+2}{2x-3}\]
\[k(x)=4x^9-3x^4+x\]
\[p(x)=x^4\]
Again, note that algebraic functions include only the operations: \(+,-,x,/\), integer exponents and rational exponents.
Non-Algebraic Functions
So, if an algebraic function can include only the operations \(+, -, x, /\), exponents (and roots), then what about all those other functions that include other operators, or have the variable at different points than the ones discussed? Those are all non-algebraic functions!
Some examples of non-algebraic functions are:
Trigonometric functions, like: \(f(x)=\sin(2x)\)
Hyperbolic functions, like : \(f(x)=\cosh(x)\)
Exponential functions, like \(f(x)=e^x\)
Logarithmic functions, like : \(f(x)=ln(x)\)
Absolute value functions, like : \(f(x)=|x|\)
Differential Calculus with Algebraic Functions
When it comes to graphing algebraic functions, sometimes we need to take Derivatives and Higher-Order Derivatives to find the critical points and inflection points of the graphs of these functions.
Derivatives of Algebraic Functions – Critical Points
If we want to find the critical points on the graph of an algebraic function, we must know how to take the derivative of the function. This can be found in our article on Derivatives.
What is a critical point?
Let c be an interior point in the domain of \(f(x)\). We say that \(x=c\) is a critical point of the function \(f(x)\) if:
- \(f(x)\)exists and if either:
- \(f'(c)=0\) or
- \(f'(c)\) doesn't exist
f(c) MUST EXIST for \(x=c\) to be a critical point!
But what is a critical point, exactly?
Wherever we have a critical point of a function, there is either a horizontal tangent, a vertical tangent, a sharp turn, or a change in concavity at that point.
- If the critical point is \(f'(c)=0\), we have a horizontal tangent at that point. This means that either:
- The function has a maximum or minimum point at \( c \).
- If we have \( f(x)=x^{2} \), then \( f'(0)=0 \) and \( x=0 \) is a critical point. At this critical point, we have a horizontal tangent and an absolute minimum at \( x=0 \).
- The function has a maximum or minimum point at \( c \).
Fig. 2. If we have \( f(x)=x^{2} \), then \( f'(0)=0 \) and \( x=0 \) is a critical point in pink. At this critical point, we have a horizontal tangent in green and an absolute minimum at \( x=0 \).
- Or the function has an inflection point at \( c \).
- If we have \( f(x)=x^{3} \), then \( f'(0)=0 \) and \( x=0 \) is a critical point. At this critical point, we have a horizontal tangent and a change in concavity, but no minimum or maximum.
Fig 3. If we have \( f(x)=x^{3} \), then \( f'(0)=0 \) and \( x=0 \) is a critical point in pink. At this critical point, we have a horizontal tangent in green and a change in concavity, but no minimum or maximum.
- If the critical point is \(f'(c)=\nexists\) (\(\nexists\) means does not exist), we have either:
- A vertical tangent atpoint \( c \).
- If we have \( f(x)=x^{\frac{1}{3}} \), then \( f'(0)=DNE \) and \( x=0 \) is a critical point. At this critical point, we have a vertical tangent and a change in concavity.
- A vertical tangent atpoint \( c \).
Fig. 4. If we have \( f(x)=x^{\frac{1}{3}} \), then \( f'(0)=DNE \) and \( x=0 \) is a critical point in pink. At this critical point, we have a vertical tangent in green and a change in concavity.
- Or a sharp turn at point \( c \).
- If we have \( f(x)=\sqrt[3]{x^{2}} \), then \( f'(0)=DNE \) and \( x=0 \) is a critical point. At this critical point, we have a sharp turn and a change in concavity.
Fig. 5. If we have \( f(x)=\sqrt[3]{x^{2}} \), then \( f'(0)=DNE \) and \( x=0 \) is a critical point in pink. At this critical point, we have a sharp turn and a change in concavity.
To find a critical point of a function, we follow these steps:
Find the Derivative of the function.
Set the derivative equal to 0.
Solve for \(x\), if possible.
Find all values of \(x\) (if any) where the derivative is undefined.
All the values of \(x\) (if they are within the domain of the original function) found in steps 2 and 3 are the \(x\)-coordinates of the critical points.
Find the \(y\)-values of the critical points by plugging the \(x\)-values into the original function and solving for the \(y\)-coordinate.
Higher-Order Derivatives of Algebraic Functions – Inflection Points
If we want to find the inflection points on the graph of an algebraic function, we must know how to take the second derivative of the function. This can be found in our article on Higher-Order Derivatives.
What is an inflection point?
If \(f(x)\) is continuous at \(a\) and \(f(x)\) changes concavity at \(a\), then the point \((a, f(a))\) is an inflection point of \(f(x)\).
But what is an inflection point, exactly?
Wherever we have an inflection point on a function, that is where it changes from either:
- Convex to Concave, or
- Concave to Convex
And what do Concave and Convex mean? 1
Convex (also called Concave Upward or Convex Downward) is when the slope of a function is increasing.
Concave (also called Concave Downward or Convex Upward) is when the slope of a function is decreasing.
Note that the AP Calculus Exams (and likely your AP Calculus teacher) use the terms Concave Up and Concave Down.
Fig. 6. The difference between concave up in blue and concave down in green.
So, how do we find where a function changes concavity?
We find the second derivative of the function.
If the second derivative of a function is positive, then it is Convex.
If the second derivative of a function is negative, then it is Concave.
Algebraic Functions Graphs
The graphs of all these types of algebraic functions vary widely from each other. However, the general procedure to graph an algebraic function is as follows:
Calculate the x-intercepts by setting and solving for \(x\).
Calculate the y-intercepts by setting and solving for \(y\).
Find and plot any asymptotes.
Find and plot any critical points.
Find and plot any inflection points.
Calculate some extra points along the curve until you get a good feel for how the curve looks.
Plot all these points and draw the curve that connects them.
Graph the function:
\[f(x)=\dfrac{1}{x}\]
Solution:
Calculate the x-intercepts by setting \(y=0\) and solving for \(x\).
\(f(x)=y=\dfrac{1}{x} \rightarrow f(x)\) can be replaced with \(y\).
\(0=\dfrac{1}{x}\) set \(y=0\) and cross-multiply to solve for \(x\).
\(0 \neq 1 \rightarrow \)There are no \(x\)-intercepts.
Calculate the \(y\)-intercepts by setting \(x=0\) and solving for \(y\).
\(f(x)=y=\dfrac{1}{x}=f(x)\) can be replaced with \(y\).
\(y=\dfrac{1}{0}\) set \(x=0\) and solve for \(y\).
\(y \neq \dfrac{1}{0}=\infty\) you can't divide by \(0\)! There are no \(y\)-intercepts.
Find and plot any asymptotes.
Since the expression \(\dfrac{1}{x} \rightarrow 0\) as \(\rightarrow \pm \infty\), there is a horizontal asymptote at the line \(y=0\) (the \(x\)-axis).
Since the expression \(\dfrac{1}{x} \rightarrow \pm \infty \) as \(x \rightarrow 0\), there is a vertical asymptote at the line \(x=0\) (the \(y\)-axis).
Find and plot any critical points.
Find the derivative of the function (check out our article on Derivatives for more info):
\(\dfrac{d}{dx}\dfrac{1}{x}=-\dfrac{1}{x^2}\)
Set the derivative equal to \(0\), and solve for \(x\).
\(0=-\dfrac{1}{x^2}\)
\(0 \neq -1\) while the point where \(x=0\) is a possible candidate for a critical point because \(f'(0)=\nexists\), there are no critical points because we proved earlier that \(f(0)\) does not exist.
Find and plot any inflection points.
Find the second derivative of the function (check out our article on Higher-Order Derivatives for more info):
\(\dfrac{d^2}{dx^2}\dfrac{1}{x}=\dfrac{2}{x^3}\)
Set the second derivative equal to \(0\), and solve for \(x\).
\(0=\dfrac{2}{x^3}\)
\(0 \neq 2\)the second derivative is never 0, and it is undefined when \(x=0\). There are no inflection points because the original function is not defined at \(0\).
Calculate some extra points along the curve until you get a good feel for how the curve looks.
\(x\) \(1/x\) \(-4\) \(-1/4\) \(-3\) \(-1/3\) \(-2\) \(-1/2\) \(-1\) \(-1\) \(0\) undefined \(1\) \(1\) \(2\) \(1/2\) \(3\) \(1/3\) \(4\) \(1/4\)
Plot all these points and draw the curve that connects them.
Fig. 7. Plotted points.
Finding Domain and Range of Algebraic Functions
Finding the domain and range of algebraic functions depends on the type of algebraic function we are considering.
For polynomial functions:
The domain for all polynomial functions is all real numbers: \(-\infty, \infty\).
The range for polynomial functions depends on both the order of the polynomial and the \(y\)-values of the graph.
If the order of the polynomial is odd, then the range is always \(-\infty, \infty\).
If the order of the polynomial is even, then the range depends on the minimum and/or maximum \(y\)-value(s).
Find the domain and range of the function:
\[f(x)=x^2+4\]
Solution:
Since this is a polynomial function:
- The domain is all real numbers: \(-\infty, \infty\).
To find the range, we recognize that this is the equation of a parabola. We can rewrite the equation in vertex form as:
\[y=a(x-h)^2+k\]
\[y=(x-0)^2+4\]
Where,
- \(a\) is the leading coefficient
- \((h, k)\) is the vertex of the parabola
Since the leading coefficient is positive, we know the parabola opens upward. This means the vertex is the lowest point of the parabola.
Therefore, the range is \([4, \infty)\) .
We can graph the function to double-check our work:
Fig. 8. The graph of a polynomial function.
For rational functions:
Finding the domain of rational functions requires that we follow the rule that the denominator cannot equal \(0\).
Finding the range of rational functions requires that we:
First, solve the function for \(x\)
Then, apply the rule that the denominator cannot equal \(0\).
Note that if the polynomials within the rational function have orders higher than \(2\), this process quickly becomes difficult. To find the range of rational functions whose orders are \(3\) or higher, we would need more differential analysis.
Find the domain and range of the function:
\[f(x)=\dfrac{3x-1}{5x+2}\]
Solution:
Since this is a rational function, we must follow the rule that the denominator cannot equal \(0\). Therefore, the domain is the set of all real numbers except where the denominator equals \(0\).
- Domain: \((-\infty, -2/5) \cup (-2/5, \infty)\).
To find the range, we need to find the values of \(y\) for which there exists a real number of \(x\) such that:
\[y=\dfrac{3x-1}{5x+2}\]
- Multiply both sides of the equation by \(5x+2\).
- \(5xy+2y=3x-1\)
- Move all terms with an x to the right of the equals sign. Move all terms without an x to the left of the equals sign.
- \(2y+1=3x-5xy\)
- Factor out the x on the right side of the equation.
- \(2y+1=x(3-5y)\)
- Solve for \(x\).
- \(x=\dfrac{2y+1}{3-5y}\)
- If \(y=3/5\), the equation has no solution because that would set the denominator equal to \(0\). Therefore, the range is:
- Range: \(\left( -\infty, \dfrac{3}{5} \right) \cup \left( \dfrac{3}{5}, \infty \right) \).
We can graph the function to double-check our work:
Fig. 9. The graph of a rational function.
For power/root functions:
Both domain and range of power and root functions depend on the exponent. We need to analyze each function on a case-by-case basis, as the domain and range of these functions are pretty “sensitive” to changes in exponents.
Find the domain and range of the function:
\[f(x)=\sqrt{4-x^2}\]
Solution:
To find the domain:
- Since this is a square root function, we need to keep what is inside the radical greater than or equal to 0.
- So, \(4-x^2 \geq 0\).
- Factor what is under the radical.
- \(4-x^2=(2-x)(2+x) \geq 0\) this inequality holds only if both terms are positive or both terms are negative.
- For both terms positive:
- \(2-x \geq 0\) and \(2+x \geq 0\)
- \(2 \geq x\) and \(x \geq -2\)
- Therefore, \([-2, 2]\) must be part of the domain.
- For both terms negative:
- \(2-x \leq 0\) and \(2+x \geq 0\)
- \(2 leq x\) and \(x \geq -2\)
- There are no values of \(x\) that can satisfy both of these inequalities.
- For both terms positive:
- \(4-x^2=(2-x)(2+x) \geq 0\) this inequality holds only if both terms are positive or both terms are negative.
- Domain: \([-2,2] \rightarrow -2 \leq x \leq 2\)
To find the range:
- We use the domain to find the range:
- If the domain is \(-2 \leq x \leq 2\), then \(0 \leq 4-x^2 \leq 4 \).
- Therefore, \(0 \leq \sqrt{4-x^2} \leq 2\).
- Range: \([0, 2] \rightarrow 0 \leq y \leq 2 \).
We can graph the function to double-check our work:
Fig. 10. The graph of a root function.
Algebraic Functions Example Problems
Which of the following are algebraic functions?
\[1.- f(x)=\sin(x)\]
\[2.- f(x)=x^2-x\]
\[3.- f(x)=ln(x)\]
\[4.- f(x)=|x-2|\]
\[5.- f(x)=x+9\]
\[6.- f(x)=\dfrac{1}{x-5}\]
\[7.- f(x)=x^{\dfrac{2}{5}}\]
\[8.- f(x)=tanh(x)\]
\[9.- f(x)=e^x\]
\[10.- f(x)=x^4-2x-6\]
\[11.- f(x)=\dfrac{2x^5-6x}{x^3}\]
Solutions:
- This is NOT an algebraic function.
- This is an algebraic function.
- This is NOT an algebraic function.
- This is NOT an algebraic function.
- This is an algebraic function.
- This is an algebraic function.
- This is an algebraic function.
- This is an algebraic function.
- This is NOT an algebraic function.
- This is NOT an algebraic function.
- This is an algebraic function.
Find the domain of each of the following:
\[f(x)=\dfrac{3}{x^2-1}\]
\[f(x)=\dfrac{2x+5}{3x^2+4}\]
\[f(x)=\sqrt{4-3x}\]
\[f(x)=\sqrt[3]{2x-1}\]
Solutions:
- We can't divide by \(0\), so the domain is the set of values of \(x\) such that \(x^2-1 \neq 0\).
- Domain: \((-\infty, -1) \cup (-1, 1) \cup (1, \infty) \).
- Again, we can't divide by \(0\), so we need to determine the values of x where the denominator is \(0\).
- \(3x^2+4>0\) for all real numbers of \(x\).
- Domain: \(-\infty, \infty \)
- \(3x^2+4>0\) for all real numbers of \(x\).
- Since the square root of a negative number is not a real number, we must determine the values of \(x\) that satisfy \(4-3x \geq 0\) .
- \(3x \leq 4\)
- \(x \leq \dfrac{4}{3}\)
- Domain: \((-\infty, 4/3]\)
- Here, the cube root is defined for all real numbers.
- Domain: \(-\infty, \infty\)
Algebraic Functions – Key takeaways
- Algebraic functions include only the algebraic operations:
- Addition: \(+\)
- Subtraction: \(-\)
- Multiplication: \(x\)
- Division: \(/\)
- Powers (aka exponents): \(x^c\), where \(c\) is any real number
- Roots: \((\sqrt[n]{x})^m=x^{m/n}\), where \(m\) is any real number and \(n\) is a positive integer greater than \(1\).
- The types of algebraic functions are:
- Polynomial functions
- Rational functions
- Power/Root functions
- Any function that contains trigonometric functions, hyperbolic functions, logarithms, variables in the power (or exponent), or an absolute value is not an algebraic function.
- Related reading:
- Functions
- Polynomial Functions
- Derivatives
- Higher-Order Derivatives
References
- https://math.stackexchange.com/questions/2364116/how-to-remember-which-function-is-concave-and-which-one-is-convex/2364123#2364123
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Frequently Asked Questions about Algebraic Functions
What is an algebraic function?
An algebraic function is a function that involves only the algebraic operations: addition, subtraction, multiplication, division, rational powers, and roots.
What is an example of an algebraic function?
Two examples of algebraic functions are rational functions and root functions.
What are the types of algebraic functions?
The types of algebraic functions are:
- Polynomial Functions
- Rational Functions
- Power/Root Functions
How do you solve algebraic functions?
Solving algebraic functions usually involves finding their domains and/or ranges and graphing them.
Why are algebraic functions important?
Learning algebraic functions are important for several reasons:
- Helps you develop problem-solving skills
- Learn how to structure and organize problems
- Learn how to approach problems that have no solution
- Learn to think abstractly
- Learn to visualize situations
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