Chapter 1.5 Manipulating Function
The functions we will encounter are very complicated, but the principle of decomposition can help with understanding the functions. In this section, we will cover various ways in which decomposition can help us study a function.
Restriction to Subdomains
First, the function we study may have a natural domain that is not ideal to consider. For example \(f(x)=x^2\) has a natural domain and domain of all real numbers. However, if \(x\) describes time, then this domain doesn’t make physical sense. So, we will talk about restricting a function to a set.
Restriction of a Function to a Set
Take \(f\) to be a function with domain \(\mathcal D(f)\) and let \(A\) be a subset of \(\mathcal D(f)\).
Define the restriction of \(f\) to \(A\), denoted by \(f\big|_A\) (read: ``\(f\) restricted to \(A\)’’), to be the function where
- the domain of \(f\big|_A\) is \(A\);
- for each \(x\) in \(A\), \(f\big|_A(x) = f(x).\)
Here is a visual that describes restriction:
Practice sketching restrictions with this next example.
Example 1
Sketch \(f\) and its restriction to
- \([0,\infty),\)
- \([-2, 0) \cup (1, 3]\),
where \(f\) is the line given by \[f(x) = 2x+1.\]
The function \(f\) is a line. On all of \(\mathbb{R}\) it looks like this:
On \([0,\infty)\) it looks like this
On \([-2,0)\cup(1,3]\) it looks like this
Now, let us look at restricting \({\rm pow}_2\).
Example 2
Sketch \(f\) and its restriction to
- \([0,\infty)\) and
- \([-1, 0] \cup (1, 2]\),
where \(f\) is given by \[f(x) = x^2.\]
The function \(f\) on all of \(\mathbb{R}\) it looks like this:
The function \(f\) on all of \([0,\infty)\) it looks like this:
The function \(f\) on all of \([0,\infty)\) it looks like this:
The Algebra of Functions
Just as we can create new sets by using set operations, we can create new function by using certain operations.
First, functions can be created by summing up two functions or multiplying two functions.
Sum and Product of Functions
Assumption: \(\mathcal D(f) \cap\mathcal D(g)\) is not empty.
Define the functions \(f+g\) and \(f\cdot g\) by \[(f+g)(x) = f(x) + g(x) \quad {\rm and}\quad (f\cdot g)(x) = f(x) \cdot g(x).\]
These new functions are defined only on the smaller domain \(D\) with \[D = \mathcal D(f)\cap \mathcal D(g).\]
Do you see why the domain of these new functions must be \(\mathcal{D}(f)\cap \mathcal{D}(g)?\)
Use the next example to make sense of that question.
Example 3
Take \(f\) to be the function that is defined for all \(x\) in \((-3, 5]\) by \[f(x) = x^2.\] Take \(g\) to be the function that is defined for all \(x\) in \([-1, 7)\) by \[g(x) = x-3.\] Can you evaluate \(f+g\) and \(f\cdot g\) at the points \(-2\), \(-1\), and \(3\)? If so, evaluate the functions at these points.
Before evaluating, first figure out the common domain.
The domain of \(f\), \(\mathcal{D}(f)\) is \((-3,5]\).
The domain of \(g\), \(\mathcal{D}(g)\) is \([-1,7)\).
The domain of \(f+g\) and \(f\cdot g\) is \[D=\mathcal{D}(f)\cap\mathcal{D}(g)=[-1,5].\]
The functions \(f+g\) and \(f\cdot g\) are both defined at \(-1\) and \(3\) because it is in the interval \(D.\) Here are the values of \(f+g\) and \(f\cdot g\):
- \[(f+g)(-1)=f(-1)+g(-1)=1+(-4)=-3,\]
- \[(f+g)(3)=f(3)+g(3)=9+0=9,\]
- \[(f\cdot g)(-1)=f(-1)\cdot g(-1)=1\cdot(-4)=-4,\]
- \[(f\cdot g)(3)=f(3)\cdot g(3)=9\cdot 0=0.\]
The functions \(f+g\) and \(f\cdot g\) are not defined at \(-2\) because it is not in the common domain \[D.\]
One example of a function that is sum of certain kinds of functions is a polynomial.
Polynomial
A polynomial is a finite sum of monomials. Assume \(k\) and \(n\) are natural numbers.
A term of the form \(a_kx^k\) is called a degree \(k\) term and the number \(a_k\) is the coefficient of the degree \(k\) term.
The largest \(n\) so that \(a_n\) is not zero is called the degree of \(f\) and is denoted by \(\deg(f)\). The number \(a_n\) is called the leading coefficient and the term \(a_nx^n\) is the leading term.
Non-zero constant polynomials are called degree zero polynomials.
Use the next example to identify these key features.
Example 4
Take \(f\) to be the function that is defined by \[f(x) = 4x^3 + 3x^2 + x - 6.\] Evaluate \(f\) at \(-2\), \(0\), and \(1\). What is the degree of \(f\)? What is the coefficient of the degree 2 term?
The evaluation of \(f\) at \(-2\), \(0\), and \(1\) is given below:
- \[f(-2)=4\cdot(-2)^3+3\cdot(-2)^2+(-2)-6=-28\]
- \[f(0)=4\cdot(0)^3+3\cdot(0)^2+(0)-6=-6\]
- \[f(1)=4\cdot(1)^3+3\cdot(1)^2+(1)-6=2\]
- \[(f\cdot g)(3)=f(3)\cdot g(3)=9\cdot 0=0.\]
The degree of \(f\) is \(3\) and the coefficient of the degree 2 term is \(3\).
Besides summing and multiplying, dividing two functions also produces a new function.
Quotient of Functions
Suppose that \(\mathcal D(f)\cap \{x\in \mathcal D(g)\colon g(x) \ne 0\}\) is not empty.
For each \(x\) in \(\mathcal D(f)\cap \{x\in \mathcal D(g)\colon g(x) \ne 0\}\), define the function \(\frac{f}{g}\) by \[\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}.\]
Here are two examples.
Example 5
Determine \(\mathcal D\!\left(\frac{f}{g}\right)\), where \(f\) and \(g\) are the functions given by \[f(x) = 2x+1 \quad {\rm and} \quad g(x) = x-5.\]
Find the domain of the numerator, \(f\) and the denominator, \(g\).
In this example, both \(f\) and \(g\) are linear functions, so
- \[\mathcal{D}(f)=(-\infty,\infty)\]
- \[\mathcal{D}(g)=(-\infty,\infty)\]
Determine the zero set of the denominator \(g\): \[g(x)=0\] whenever \[x-5=0.\] Thus \(x=5\) is the value we exclude from the domain of \(g\):
\[\{x\in \mathcal D(g)\colon g(x) \ne 0\}=(-\infty,5)\cup(5,\infty),\]
Thus the domain \(\mathcal D\!\left(\frac{f}{g}\right)\) is \[\mathcal D(f)\cap \{x\in \mathcal D(g)\colon g(x) \ne 0\}=(-\infty,\infty)\cap\left[(-\infty,5)\cup(5,\infty)\right]=(-\infty,5)\cup(5,\infty).\]
Try this next example as well.
Example 6
Determine \(\mathcal D\!\left(\frac{f}{g}\right)\), where \(f\) and \(g\) are the functions given by \[f(x) = x \quad {\rm and} \quad g(x) = x.\]
Find the domain of the numerator, \(f\) and the denominator, \(g\).
In this example, both \(f\) and \(g\) are linear functions, so
- \[\mathcal{D}(f)=(-\infty,\infty)\]
- \[\mathcal{D}(g)=(-\infty,\infty)\]
Determine the zero set of the denominator \(g\): \[g(x)=0\] whenever \[x=0.\] Thus \(x=0\) is the value we exclude from the domain of \(g\):
\[\{x\in \mathcal D(g)\colon g(x) \ne 0\}=(-\infty,0)\cup(0,\infty),\]
Thus the domain \(\mathcal D\!\left(\frac{f}{g}\right)\) is \[\mathcal D(f)\cap \{x\in \mathcal D(g)\colon g(x) \ne 0\}=(-\infty,\infty)\cap\left[(-\infty,0)\cup(0,\infty)\right]=(-\infty,0)\cup(0,\infty).\]
Another operation that produces a new function is called composition.
Composite Function
Take \(f\) and \(g\) to be real valued functions.
For every \(x\) in \(\mathcal D(g)\) such that \(g(x)\) is in \(\mathcal D(f)\), define the function \(f\circ g\) (read: ``\(f\) composed with \(g\)’’) by \[(f\circ g)(x) = f(g(x)).\]
The function \(f\circ g\) is a composite function.
The way this operation works is that instead of evaluating the function \(f\) at \(x\), evaluate the function \(f\) at \(g(x).\)
To better understand this, here is an example of a composition.
Example 7
Find a formula for the function \(f\circ g\) and for the function \(g\circ f\), where \(f\) and \(g\) are given by \[f(x) = x^2 + 1 \quad {\rm and}\quad g(x) = 2x+3.\]
The function \(f(x)=x^2+1\) can be broken up as \(f(\star)=(\star)^2+1\). In words, square the input then add one.
The function \(g(x)=2x+1\) can be broken up as \(f(\star)=2(\star)+3\). In words, multiply the input by 2 then add three.
The function \(f\circ g\) is defined as \((f\circ g)(x)=f(g(x)).\) Therefore
\[\begin{align*} (f\circ g)(x)&=f(g(x))\\ &=(g(x))^2+1\\ &=(2x+3)^2+1&&\text{ recall } \star^2=\star\cdot\star \text{ so }(2x+3)^2=(2x+3)(2x+3)\\ &=(4x^2+12+9)+1\\ &=4x^2+12x+10 \end{align*}.\]
The function \(g\circ f\) is defined as \((g\circ f)(x)=g(f(x)).\) Therefore
\[\begin{align*} (g\circ f)(x)&=g(f(x))\\ &=2\cdot(f(x))+3\\ &=2\cdot(x^2+1)+3\\ &=2x^2+2+3\\ &=2x^2+5 \end{align*}.\]
We will use the principle of decomposition to determine the domain of a function.
Example 8
Take \(f\) and \(g\) to be the functions with \[\mathcal D(f) = (-4, 12]\quad {\rm and} \quad \mathcal D(g) = (-\infty, 10).\] If \(\{-5, 3, 7\}\) is the zero set of \(g\), then what are the domains of \(f+g\), \(f\cdot g\), and \(\frac{f}{g}\)?
The intersection \(\mathcal D(f)\cap \mathcal D(g)\) is \((-4,10)\).
Therefore, the domain of \(f+g\) and \(f\cdot g\) is \((-4,10).\)
The domain of \(\frac{f}{g}\) is \(\mathcal{D}(f)\cap\{x\in \mathcal D(g)\colon g(x) \ne 0\}\). The set \(\{x\in \mathcal D(g)\colon g(x) \ne 0\}\) is the domain of \(g\) with the zero set of \(g\) removed. So \[\{x\in \mathcal D(g)\colon g(x) \ne 0\}=(-\infty,10)\setminus\{-5,3,7\}=(-\infty,-5)\cup(-5,3)\cup(3,7)\cup(7,10).\]
Therefore
\[\mathcal{D}(f)\cap\{x\in \mathcal D(g)\colon g(x) \ne 0\}=(-4,12]\cap\left((-\infty,-5)\cup(-5,3)\cup(3,7)\cup(7,10)\right)=(-4,3)\cup(3,7)\cup(7,10).\]
In this next example, practice determining the domain of functions with given formulas.
Example 9
Take \(f\) and \(g\) to be given by \[f(x) = \sqrt{x -4}\quad \text{and}\quad g(x) = \sqrt{9-x}.\] What are the domains of the two functions, their product, and the quotient \(\frac{f}{g}\)?
The function \(\mathrm{pow}_{1/2}(\star)=\sqrt{\star}\) is defined as long as \(\star\geq 0\).
The function \(f(x)=\sqrt{x-4}\) is defined as long as \(x-4\geq 0\) which is equivalent to \(x\geq 4\).
So \(\mathcal D(f)=[4,\infty)\).
The function \(g(x)=\sqrt{9-x}\) is defined as long as \(9-x\geq 0\) which is equivalent to \(9\geq x\) or \(x\leq 9\).
So \(\mathcal D(g)=(-\infty,9]\).
Now calculate \(\mathcal D(f)\cap \mathcal{D}(g).\) The intersection is \([4,9]\) and thus the domain of \((f\cdot g)\) is \([4,9]\).
To calculate \(\mathcal D\left(\frac{f}{g}\right)=\mathcal{D}(f)\cap\{x\in \mathcal D(g)\colon g(x) \ne 0\}\), find \(\{x\in \mathcal D(g)\colon g(x) \ne 0\}\), which is the domain of the denominator \(g\) (\((-\infty,9]\)) with the zero set of \(g\) removed. The function \(g(x)=\sqrt{9-x}\) is zero when \(9-x=0\); the zero set is \(\{9\}\). Remove \(9\) from the set \((-\infty,9].\)
Thus \(\{x\in \mathcal D(g)\colon g(x) \ne 0\}=(-\infty,9).\)
Our final answer for \(\mathcal D\left(\frac{f}{g}\right)\) is \(\mathcal D(f)=[4,\infty)\) intersect \(\{x\in \mathcal D(g)\colon g(x) \ne 0\}=(-\infty,9)\), which simplifies to \([4,9).\)
In this example, practice finding the domain of a composite function.
Example 10
Suppose that \(f\) and \(g\) are given by \[f(x) = \frac{1}{x} \quad \text{and}\quad g(x) = \sqrt{x - 4}.\]
What is the domain of \(f\circ g\)?
What is the domain of \(g\circ f\)?
Does \(f\circ g = g\circ f\)?
Identify the domain of \(f\) and \(g\).
The domain of \(f(x)=\frac{1}{x}\) is \((-\infty,0)\cup(0,\infty).\)
The domain of \(g(x)=\sqrt{x-4}\) is \([4,\infty).\)
To find domain of \(f\circ g\), first find domain of \(g\). It is \([4,\infty).\)
The domain of \(f\circ g\) is all \(x\) that are in domain of \(g\) and for which the output \(g(x)\) is in the domain of \(\mathcal{D}(f).\)
In this example, the output \(g(x)=\sqrt{x-4}\) must be a real number in \(\mathcal{D}(f)=(-\infty,0)\cup(0,\infty).\) Thus \(\sqrt{x-4}\) cannot equal \(0\). This happens when \(x=4\).
So the answer is \([4,\infty)\) remove \(x=4.\)
Thus \(\mathcal{D}(f\circ g)=(4,\infty)\)
To find domain of \(g\circ f\), first find domain of \(f\). It is \((-\infty,0)\cup(0,\infty)\)
The domain of \(g\circ f\) is all \(x\) that are in domain of \(f\) and for which the output \(f(x)\) is in the domain of \(\mathcal{D}(g).\)
In this example, the output \(f(x)=\frac{1}{x}\) must be a real number in \(\mathcal{D}(g)=[4,\infty).\) Thus \(\frac{1}{x}\) must be greater than or equal to \(4\). This happens when \(x\) is in the interval \((0,\frac{1}{4}]\).
So the answer is \((-\infty,0)\cup(0,\infty)\) intersect \((0,\frac{1}{4}]\).
Thus \(\mathcal{D}(g\circ f)=(0,\frac{1}{4}]\)
The composite functions \((f\circ g)(x)=f(g(x))=\frac{1}{g(x)}\) and \((g\circ f)(x)=g(f(x))=\sqrt{f(x)-4}=\sqrt{\frac{1}{x}-4}\) are not the same.
Complicated functions can be created by using some combination of addition, multiplication, division, and the composition operation on elementary functions.
The arithmetical properties of addition and multiplication give similar properties for the associated operations, for example, \(+\) and \(\cdot\) are associative, commutative, and the operations distribute.
Be careful, however when dealing with composition. Composition is associative, but does not in general commute or distribute over addition or multiplication.
Understand this warning by carefully answering this next example.
Example 11
Take \(f\), \(g\), and \(h\) to be the functions given by \[f(x) = \frac{1}{x},\quad g(x) = 3x,\quad {\rm and}\quad h(x) = x^2.\] Calculate \(f\circ(g+h)\), \((f\circ g) + (f\circ h)\), \((g+h)\circ f\), and \((g\circ f) + (h\circ f)\).
The function \(f(x)=\frac{1}{x}\) can be broken up as \(f(\star)=\frac{1}{\star}\). In words, it is one divided by the input.
The function \(g(x)=3x\) can be broken up as \(f(\star)=3(\star)\). In words, multiply the input by 3.
The function \(h(x)=x^2\) can be broken up as \(f(\star)=(\star)^2\). In words, multiply the input by 3.
The function \(g+h\) is defined as \((g+h)(x)=3x+x^2.\)
Since \(f(\star)=\frac{1}{\star}\), we get
\[\begin{align*} f\circ(g+h)(x)&=\frac{1}{(g+h)(x)}\\ &=\frac{1}{3x+x^2}. \end{align*}\]
The function \(f\circ g\) is defined as \((f\circ g)(x)=\frac{1}{3x}.\)
The function \(f\circ h\) is defined as \((f\circ h)(x)=\frac{1}{x^2}.\)
Therefore
\[\begin{align*} (f\circ g)(x)+(f\circ h)(x)&=f(g(x))+f(h(x))\\ &=\frac{1}{3x}+\frac{1}{x^2}\\ &=\frac{x+3}{3x^2}. \end{align*}\]
For \((g+h)\circ f\), and \((g\circ f) + (h\circ f)\). we have
\[\begin{align*} (g+h)(x)\circ f(x)&=3(f(x))+(f(x))^2\\ &=3\frac{1}{x}+\left(\frac{1}{x}\right)^2\\ &=\frac{3}{x}+\frac{1}{x^2}. \end{align*}\]
\[\begin{align*} (g\circ f)(x)+(h\circ f)(x)&=3(f(x))+(f(x))^2\\ &=3\frac{1}{x}+\left(\frac{1}{x}\right)^2\\ &=\frac{3}{x}+\frac{1}{x^2}. \end{align*}\]
We will use the power of decomposition to handle the following problem.
Example 12
Take \(a\), \(b\), \(c\), \(d\), and \(e\) to be the functions given by \[a(x) = x,\; b(x) = x^2,\; c(x) = x+1,\; d(x) = 5x,\; e(x) = \sqrt{x}.\] For each \(x\) where the function \(f\) is defined, calculate \(f(x)\), where \[f = \frac{a\cdot e}{c\circ b\circ c} + e\circ e\circ(3a + c\circ d\circ b).\]
Use decomposition to split \(f\): \[f(x)=\text{part}_1(x)+\text{part}_2(x).\]
For \(\text{part}_1(x)\)
\[\text{part}_1=\frac{a\cdot e}{c\circ b\circ c}=\frac{\text{Part A}}{\text{Part B}}.\]
For \(\text{Part A}\) and \(\text{Part B}\)
\[\text{Part A}(x)= a(x)\cdot e(x)=x \sqrt{x}\],
\[\begin{align}\text{Part B}(x)&= (c\circ b\circ c)(x)\\ &=c\circ \left[b(c(x))\right]\\ &=c[(x+1)^2]\\ &=c[x^2+2x+1]\\ &=(x^2+2x+1)+1\\ &=x^2+2x+2.\\ \end{align}\]
Thus
\[\begin{align} \text{part}_1(x)=&\frac{(a\cdot e)(x)}{(c\circ b\circ c)(x)}\\ &=\frac{\text{Part A}(x)}{\text{Part B}(x)}\\ &=\frac{x\sqrt{x}}{x^2+2x+2}. \end{align}\]
For \(\text{part}_2(x)\)
\[\text{part}_2=(e\circ e)\circ(3a+c\circ d\circ b)=\text{Part C}\circ\text{Part D}.\]
For \(\text{Part C}\) and \(\text{Part D}\)
\[\text{Part C}(x)=(e\circ e)(x)=e(\sqrt{x})=\sqrt{\sqrt{x}}\],
\[\begin{align}\text{Part D}(x)&= 3a(x)+c\circ [d(b(x))]\\ &=3x+c[5(x)^2]\\ &=3x+[5x^2]+1\\ &=5x^2+3x+1 \end{align}\]
Thus
\[\begin{align} \text{part}_2(x)=&\text{Part C}\circ\text{Part D}(x)\\ &=\sqrt{\sqrt{\text{Part D}(x)}}\\ &=\sqrt{\sqrt{5x^2+3x+1}}. \end{align}\]
Altogether,
\[f(x)=\frac{x\sqrt{x}}{x^2+2x+2}+\sqrt{\sqrt{5x^2+3x+1}}.\]
Decomposing Functions
A function \(h\) may be built as sums, products, and composites of other simpler functions.
- Sum: \(h(x) = f(x) + g(x)\)
- Product: \(h(x) = f(x)g(x)\)
- Quotient: \(h(x) = \dfrac{f(x)}{g(x)}\)
- Composite: \(h(x) = f(g(x))\).
Understanding the structural components of a function means that we understand how a function is built from other functions.
First, let’s look at a sum.
Example 13
Write \(f\) as a sum of functions, where \[f(x) = x^2 - 3x^5.\]
Here is one way to write \(f\) as a sum.
Take \(g(x)=x^2\) and \(h(x)=-3x^2\). Then \(f(x)=(g+h)(x)\).
We can check this:
\[\begin{align} (g+h)(x)&=g(x)+h(x)\\ &=x^2-3x^2\\ &=f(x) \end{align}\]
Now, let’s look at a product.
Example 14
Write \(f\) as a product of functions, where \[f(x) = x^2\sqrt{x+5}.\]
Here is one way to write \(f\) as a product.
Take \(g(x)=x^2\) and \(h(x)=\sqrt{x+5}\). Then \(f(x)=g(x)\cdot h(x)\).
We can check this:
\[\begin{align} (g+h)(x)&=g(x)\cdot h(x)\\ &=x^2\cdot \sqrt{x+5}\\ &=x^2\sqrt{x+5} \end{align}\]
Next, a composition.
Example 15
Write \(f\) as a composite function, where \[f(x) = \sqrt{x^2 -2}.\]
Here is one way to write \(f\) as a composite.
Take \(g(x)=\sqrt{x}\) and \(h(x)=x^2-2\). Then \(f(x)=(g\circ h)(x)\).
We can check this:
\[\begin{align} (g\circ h)(x)&=g(h(x))\\ &=\sqrt{h(x)}\\ &=\sqrt{x^2-2}\\ &=f(x) \end{align}\]
Now, let’s decompose a more complicated function.
Example 16
Decompose \(f\) into sums, products, quotients, and composites of more elementary functions, where \[f(x) = \dfrac{\sqrt{x^2 + \sqrt{x+1}}}{3x+1}.\]
Here is one way to write \(f\) as a composite.
Take \(g(x)=x^2\), \(h(x)=x+1\), \(j(x)=\sqrt{x}\), and \(k(x)=3x+1\). Then the numerator, \(\sqrt{x^2+\sqrt{x+1}}\) can be broken up like this: - the inside of the square root, \(x^2+\sqrt{x+1}\), is the sum of \(g(x)=x^2\) and \(\sqrt{x+1}\). - the \(\sqrt{x+1}\) can be written as the composite of \(h\) and \(j\). Specifically \(\sqrt{x+1}=\sqrt{h(x)}=(j\circ h)(x)\) - the inside of the square root, \(x^2+\sqrt{x+1}\) is therefore equal to \(g+(j\circ h)\). - the numerator is the square root of that sum, so it can be rewritten as \(j\circ(g+j\circ h)\) - the denominator is simply \(k(x)=3x+1\)
Thus \(f(x)=\left(\frac{j\circ(g+j\circ h)}{k}\right)(x)\)
Let’s decompose an even more complicated function.
Example 17
Decompose \(f\) into sums, products, quotients, and composites of more elementary functions, where \[f(x) = (2x+1)^2 + x\sqrt{\sqrt{x^2 + \frac{\sqrt{x}}{5x^2 + 1}}}.\]
Here is one way to write \(f\) as a composite.
Take
- \(a(x)=x\)
- \(b(x)=x^2\)
- \(c(x)=x+1\)
- \(d(x)=\sqrt{x}\)
Decompose \(f\) into the following parts
- \((2x+1)=(c\circ 2a)(x)\) so \((2x+1)^2=[b\circ(c\circ 2a)](x)\)
- \(\frac{\sqrt{x}}{5x^2+1}=\frac{d}{c\circ(5b)}(x)\), so \(x^2+\frac{\sqrt{x}}{5x^2+1}=\left[b+\frac{d}{c\circ(5b)}\right](x)\)
- \(\sqrt{\sqrt{x^2+\frac{\sqrt{x}}{5x^2+1}}}=d\circ d\left(b+\frac{d}{c\circ (5b)}\right)(x)\)
- \(x\sqrt{\sqrt{x^2+\frac{\sqrt{x}}{5x^2+1}}}=a\cdot \left[d\circ d\left(b+\frac{d}{c\circ (5b)}\right)\right](x)\) Thus \(f=[b\circ(c\circ 2a)]+a\cdot \left[d\circ d\left(b+\frac{d}{c\circ (5b)}\right)\right]\)
Computing the Range of a Function
We are going to revisit the idea of a range of a function. First, we restate the range like this:
Range of Function
Suppose that \(f\) is a function from a subset \(D\) of \(\mathbb R\) to \(\mathbb R\).
A real number \(a\) is in the range of \(f\) if there is an \(x\) in \(D\) with \[f(x) = a.\]
In general, computing the range of a function is difficult. For now, we focus on simple examples.
Example 18
Suppose that \[f(x) = \frac{3x+2}{x-1}.\] What is the domain and range of \(f\)?
The domain of \(f\) is \(\mathcal{D}(f)=(-\infty,1)\cup(1,\infty).\)
A real number \(b\) is in \(\mathcal{R}(f)\) if and only if the equation \(f(x)=b\) has a solution. In this example.
\[\begin{align*} f(x)&=b\\ \frac{3x+2}{x-1}&=b\\ 3x+2&=b\cdot(x-1)\\ 3x+2&=bx-b\\ 3x-bx&=-b-2\\ x(2-b)&=-b-2\\ x&=\frac{-b-2}{3-b} \end{align*}\] The solution, \(x\), is a real number if and only if the denominator, \(3-b\) is non-zero. Thus \(b\not 3\). As long as \(b\) is not 3, \(b\) is in the range.
The range is all real numbers except 3; \(\mathcal{R}(f)=(-\infty,3)\cup(3,\infty).\)
Use decomposition to find the domain and range of \(f\).
Example 19
Take \(g\) to be the function that is defined for all \(x\) in \((-\infty, -2]\cup [3, 10)\) by \[g(x) = x+6\quad \text{and take}\quad f = {\rm pow}_{\frac{1}{2}}\circ g.\] Determine the domain and range of \(f\).
The function \(\text{pow}_{1/2}(x)=\sqrt{x}\) has a domain of \([0,\infty)\) and range of \([0,\infty).\)
The function \(f\) is the composite function \(f(x)=\sqrt{g(x)}\).
In order for a real number \(x\) to be in the domain of \(f\), we need \(x\) to be in the domain of \(g\).
The first condition is therefore \(x\in (-\infty,-2]\cup [3,10).\)
Next, since we have \(g\) inside of \(\text{pow}_{1/2}\), we need \(g\geq 0\):
\[ \begin{align*} g(x)&\geq 0\\ x+6&\geq 0\\ x&\geq -6\\ \end{align*} \]
The second condition is therefore \(x\in [-6,\infty)\).
Take the intersection of the first condition and second condition to obtain \[\mathcal{D}(f)=[-6,-2]\cup[3,10).\]
For the range, notice that \(f(x)=\sqrt{x+6}\) is strictly increasing.
The range, can be determined by the extrema of \(f\) on \([-6,-2]\) and \([3,10).\)
- \(f(-6)=0\), \(f(-2)=2\) gives us that the interval \([0,2]\) is part of the range
- \(f(3)=3\), \(f(10)=4\) gives us that the interval \([3,4]\) is part of the range
Together, we get \(\mathcal{R}(f)=[0,2]\cup [3,4].\)