Define function. Distinguish relation and function with example. Find the domain and range of f(x) = x^2 2x 8 , x R

Definition of a Function

A function $f$ from a set $A$ to a set $B$ is a rule that assigns to every element $x\in A$ exactly one element $y\in B$.

• The set $A$ is called the domain of $f$.
• The set $B$ is called the codomain of $f$.
• The set of all actual outputs $\{f(x)\mid x\in A\}$ is called the range (or image) of $f$.

Mathematically,
$$f:A⟶B,f(x)=y\text{with}y\in B,\forall x\in A.$$

Relation vs Function

Illustration

flowchart TD
  "Relation R" --> "Multiple y for same x"
  "Function f" --> "Unique y for each x"

Domain and Range of $f(x)=\sqrt{x^2-2x-8}$

1. Determine the domain

The expression under the square root must be non‑negative:

$$x^2-2x-8\ge 0.$$

Factor the quadratic:

$$x^2-2x-8=(x-4)(x+2).$$

Solve the inequality $(x-4)(x+2)\ge 0$.
Critical points: $x=-2$ and $x=4$.
Test intervals:

Include the points where the expression equals zero ($x=-2,4$) because $\sqrt{0}=0$ is defined.

$$\boxed{\text{Domain }D=(-\infty ,-2]\cup [4,\infty )}.$$

2. Find the range

For $x$ in the domain, $f(x)$ is the square root of a non‑negative number, so $f(x)\ge 0$.
We examine the minimum value of the radicand $x^2-2x-8$ on each interval.

On $(-\infty ,-2]$:
Let $g(x)=x^2-2x-8$.
Derivative $g^{\prime }(x)=2x-2$.
Critical point $x=1$ lies outside the interval, so $g(x)$ is increasing for $x>-1$ and decreasing for $x<-1$.
On $(-\infty ,-2]$ the function is decreasing as $x$ moves rightwards, so the maximum occurs at the left endpoint $-\infty$ (unbounded) and the minimum at $x=-2$:

$$g(-2)=(-2{)}^2-2(-2)-8=4+4-8=0.$$

Thus on this interval $f(x)$ ranges from $0$ to $\infty$.

On $[4,\infty )$:
Similarly, $g(x)$ is increasing for $x>1$.
Minimum at $x=4$:

$$g(4)=4^2-2(4)-8=16-8-8=0.$$

Again, $f(x)$ starts at $0$ and increases without bound.

Combining both intervals, the smallest output is $0$ (attained at $x=-2$ and $x=4$), and there is no upper bound.

$$\boxed{\text{Range }R=[0,\infty )}.$$

Final Result
Domain: $(-\infty ,-2]\cup [4,\infty )$
Range: $[0,\infty )$