# Range and Domain of a Function

If you are not familiar yet what are domains and range of a function, you better read this introduction first. This topic requires deep understanding in Math. No wonder questions regarding this topic can be seen usually in Math contests specifically MMC. With the help of your patience and determination to win this topic, I’m sure you will be able to understand this by heart.  I have provided some important tips to keep in mind so that you would know where to start. It is very important that we know how to evaluate basic inequalities. It is also a must that you are already familiar of the following topics. If you need more problems to practice, brilliant website is a perfect for you.

Important tips:

• Identify roughly how the graph of equation looks like.
• If you are looking for domain solve for y in terms of x.
• If you are looking for range solve for x in terms of y.
• If the given equation is a combination of two or more functions, the domain or range is the intersection of individual domains or range.
• take note of restrictions of basic equations
• The functions inside square root symbol must be greater than or equal to 1. In $\sqrt{U}, U\geq 0$
• Functions in denominator must be equal to zero. In $\displaystyle\frac{1}{U}, U\ne 0$
• In logarithmic equation $f(x)=\log_b U, b>1, U>0$

Sample Problem 1:

Find the range of $f(x)=2x-1$ on $\{x|-2

Solution:

{x|-2<x<3} is the interval where we will plot the given function. It’s like a Cartesian plane cut through $x\ne -2$ and $x\ne 3$ (a dotted vertical lines). The graph of the given function is supposed to be continuous, but since plane was cut, the line has two endpoints. Since the trend of the graph is increasing the right endpoint of the graph can be calculated as follows.

x=3(right endpoint)

$f(x)=2x-1=2(3)-1=\boxed{5}$

x=-2(left endpoint)

$f(x)=2x-1=2(-2)-1=\boxed{-5}$

Thus the range of equation is $(-5,5)$  or  $\boxed{\{y|-5

Sample Problem 2:

Find the range and domain of $y=\sqrt{x-2}$

Solution:

Domain: check the first restriction. $U\geq 0$

$x-2\geq 0 \longrightarrow x\geq 2$

The domain is $\boxed{[2,+\infty)}$  or $x\geq 2$

For range:

Staring from the endpoint of the graph from 2 we test some value.

Let x=2

$y=\sqrt{x-2}$

$y=\sqrt{2-2}=0$

Let x=3

$y=\sqrt{x-2}$

$y=\sqrt{3-2}=1$

Let x=6

$y=\sqrt{x-2}$

$y=\sqrt{6-2}=2$

Continuing the process, the values of y is dependent on x. If x goes to positive infinity, same goes with y. thus, the range is $\boxed{[0,+\infty)}$

Sample Problem 3:

Find the domain of $f(x)=\sqrt{3x-\sqrt{x-1}}$

Solution:

The domain of $3x$ inside the square root is $x\geq 0$. The domain of $\sqrt{x-1}$ is $x\geq 1$. Thus the real domain of $f(x)$ is the intersection of two.

$[0,+\infty)\cup [1,+\infty) \to \boxed{[1,+\infty)}$

Sample Problem 4:

Find the domain of   $f(x)=\sqrt{\displaystyle\frac{3^x-1}{2-x}}$

Solution:

$\sqrt{3^x-1}\geq 0$

Also

$\sqrt{2-x}\ge 0, \sqrt{2-x}\ne 0$

Evaluating the inequality

$\sqrt{3^x-1}\geq 0$

$x\ge 0$

Evaluating the 2nd inequality

$\sqrt{2-x}\ge 0, \sqrt{2-x}\ne 0$

$x\le 2, x \ne 2$

$x<2$

The domain of $f(x)$ is the intersection of $x<2$ and $x\ge 0$. Thus, the domain is $0\le x<2$ or in symbol $\boxed{[0,2)}$

Sample Problem 5:

Find the domain of $f(x)=\log_{x-3} (x^2-5x+4)$

Solution:

$x^2-5x+4>0, x-3>1$

$(x-4)(x-1)>0, x>4$

$x>4,$ or $x<1,$ and $x>4$

The domain is the intersection which is $\boxed{(-\infty,1)\cup (4,+\infty)}$