The domain of a function $f$ is a set of input values for which the function is defined. It is often useful to draw a sketch to help identify the domain of a particular function.
Find the domain of the function $f(x) = 4x+2$.
This function is defined for all values of $x$ as each real number $\mathbb{R}$ inputted gives a valid output. In this case, $x$ can take all real values and the domain is the set of all real numbers $\mathbb{R}$.
Find the domain of the function $f(x) =\dfrac{1}{x}$.
In this case, the denominator can never equal zero as then the function is undefined. Therefore, we must specify that $x\neq0$ and the domain is then $D=\mathbb{R}-\{0\}$; that is the set of real numbers excluding $0$.
Find the domain of the function $f(x) = \dfrac{2}{x-3}$.
Again, the denominator can never equal zero as then the function is undefined. Therefore we must specify that $x - 3 \neq 0$ $\Rightarrow x \neq 3$ and the domain is $D=\mathbb{R}-\{3\}$; that is the set of real numbers excluding $3$.
Prof. Robin Johnson finds the domain of the function $f(x) = \sqrt{x+3}$.
Prof. Robin Johnson finds the domain of the function $f(x) = \sqrt{x^2-4x+3}$.
Prof. Robin Johnson finds the domain of the function $f(x) = \dfrac{1}{x^2-3x^2+2x}$.
The range of a function is the set of outputs that correspond with a set of given inputs; the range therefore follows from the domain of a function. It is often useful to draw a sketch to help identify the range.
Find the range of the function $f(x) = 2x+3$.
As $x\rightarrow\infty$, $f(x)\rightarrow\infty$ and as $x\rightarrow -\infty$, $f(x)\rightarrow -\infty$
So $f(x)$ can take all real values.
Find the range of the function $f(x) = \dfrac{1}{2x}$.
As $x\rightarrow\infty$, $f(x)\rightarrow 0^+$ and as $x\rightarrow -\infty$, $f(x)\rightarrow 0^-$. There is no value of $x$ that gives $f(x)=0$.
So $f(x) \neq 0$.
Find the domain and range of the function $f(x) = \sqrt{x+4}$.
Before considering this example let us first consider the case where we find the domain and range of the function $g(x) = \sqrt{x}$.
If we are considering only those outputs which are in the set of real numbers $\mathbb{R}$ we cannot take the square root of a negative number so the square root must be a positive number. That is, we define the domain of this function to be $x\geq0$.
However, there is another factor we need to consider for the range of this function and for this we need to return to our definition of a function which is a mapping that takes a set of inputs and maps these to a set of outputs whereby each input has only one permissible output. By taking the $\sqrt{x}$ (one input) there are two possible outputs $+\sqrt{x}$ and $-\sqrt{x}$ since $(-x).(-x) = x^2$ and $(x).(x)=x^2$. To satisfy our definition of a function, we need to consider only the positive outputs. We can restrict the range of a function in order to ensure it is valid and satisfies our definition that a function maps one input to exactly one output. In this case, we choose to consider only those outputs where $g(x)\geq0$ hence our range is $g(x)\geq0$.
Returning to our example $f(x) = \sqrt{x+4}$.
We must have that $x+4\geq0$ $\Rightarrow x \geq -4$. So the domain is $D=\{x:x \geq -4\}=[-4,\infty]$ and the range is $f(x)\geq0$.
Prof. Robin Johnson finds the range of the function $g(z) = z^2-5z+4$.
Prof. Robin Johnson finds the range of the function $h(y) = 1+2y-y^2$.
Prof. Robin Johnson finds the range of the function $k(x) = \dfrac{1}{x-1}$.
Prof. Robin Johnson finds the range of the function $f(t) = \sqrt{t-1}$.