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Domain Of Functions

Practice Linear Inequalities Problems, learn about the domain & the range of a relation, linear function and find the domain of the expression.

Linear Inequalities

Consider a linear function where $$a>0$$$$f(x)=ax+b$$, suppose we want to know for what values of $$x$$ is the expression positive, i.e.

$$ax+b>0\rightarrow$$ this is called a linear inequality or linear equation.

To solve this consider:

 $$ax+b>0$$ Subtract $$-b$$ from both sides $$ax>-b$$ Divide by $$'a'$$ on both sides $$x>\dfrac {-b}{a}$$ Solution to linear inequality, also represented by $$x\in\left (\dfrac {-b}{a},\infty \right)$$ Similar step for solving $$ax+b<0$$

In general:

• $$x>a$$ is represented by $$x\in(a, \infty)$$
• $$x\geq a$$ is represented by $$x\in[a, \infty)$$
• $$x< a$$ is represented by $$x\in(-\infty, a)$$
• $$x\leq a$$ is represented by$$x\in(-\infty, a]$$

For representing values of $$x$$ which are between $$'a'$$ and $$'b'$$ excluding them

• $$a<x<b$$ is represented by $$x\in(a, b)\rightarrow$$ Open interval
• $$a\leq x\leq b$$is represented by $$x\in[a, b]\rightarrow$$ Closed interval

• $$a\leq x< b$$is represented by $$x\in[a, b)\rightarrow$$ Half open interval

For what values of $$x$$ is the expression  $$f(x)=2x-7$$ positive?

A $$x<1$$

B $$x>\dfrac {7}{2}$$

C $$x<-2$$

D $$x=0$$

×

$$f(x)>0$$

$$2x-7>0$$

Add 7 to both sides of inequality

$$2x-7+7>7$$

$$2x>7$$

Divide by $$2$$ on both sides

$$x>\dfrac {7}{2}$$

We can also say $$x\in\left (\dfrac {7}{2},\infty\right)$$, so $$x>\dfrac {7}{2}$$ and $$x\in\left (\dfrac {7}{2},\infty\right)$$ are two methods of representing same solution.

For what values of $$x$$ is the expression  $$f(x)=2x-7$$ positive?

A

$$x<1$$

.

B

$$x>\dfrac {7}{2}$$

C

$$x<-2$$

D

$$x=0$$

Option B is Correct

Inequality with Equation

• Consider the inequality

$$ax + b \geq0 \space or\, \space ax + b \leq 0$$

• Here, $$ax + b \geq 0$$ means finding those values of x for which this expression is positive or 0 and $$ax + b \leq 0$$ means finding those values of x for which expression is negative or 0.
• Effectively, to the solution of ax + b > 0 (already done), we have to add the solution to $$ax + b =0$$  to obtain the solution of $$ax + b \geq 0$$.

Solve: $$3x - 7 \leq 0$$.

A $$x \leq \Large {7 \over 3}$$

B $$x \geq 2$$

C $$x \leq 10$$

D $$x \geq -1$$

×

$$3x - 7 \leq 0$$

Add 7 on both sides of inequality

$$3x - 7 + 7 \leq 7$$

$$3x \leq 7$$   (Divide by 3)

$$x \leq \large {7 \over 3}$$       or       $$x \space \epsilon (-\infty, {\large {7 \over 3}}]$$

Note the closed interval

$$x = \dfrac{7}{3}$$ is a part of solution

Solve: $$3x - 7 \leq 0$$.

A

$$x \leq \Large {7 \over 3}$$

.

B

$$x \geq 2$$

C

$$x \leq 10$$

D

$$x \geq -1$$

Option A is Correct

Find the domain of $$f(x) = \dfrac {2x^2 - 5}{x^2 + x - 6}$$.

A $$\{x \space | \space x \neq -3, x \neq 2 \}$$

B $$\{x \space | \space x \neq 10, x \neq 1 \}$$

C $$\{x \space | \space x \neq 5, x \neq 7 \}$$

D $$\{x \space | \space x \neq -5, x \neq -10 \}$$

×

Domain will be all real x except those values for which denominator is 0.

So     $$x^2 + x - 6 = 0$$        ⇒ (x + 3) (x – 2) = 0

⇒ x + 3 = 0   or x – 2 = 0

⇒ x = –3   or   x = 2

So domain is   $$\{x \space | \space x \neq -3, x \neq 2 \}$$

Note that this can also be written as

$$(-\infty, -3) \cup (-3, 2) \cup (2, \infty)$$ or $$(-\infty, \infty) - \{-3, 2\}$$

Find the domain of $$f(x) = \dfrac {2x^2 - 5}{x^2 + x - 6}$$.

A

$$\{x \space | \space x \neq -3, x \neq 2 \}$$

.

B

$$\{x \space | \space x \neq 10, x \neq 1 \}$$

C

$$\{x \space | \space x \neq 5, x \neq 7 \}$$

D

$$\{x \space | \space x \neq -5, x \neq -10 \}$$

Option A is Correct

Domain of a Function

• The set A from which the function takes the value or the set of all input values 'x' is called the domain of that function.
• The set of all output values is called range of that function.

$$f(\underbrace{x}_{domain})=\underbrace{ formula \,of\, an\,expression}_{range}$$

• A domain is the collection of all values of x for which function is properly defined.
• The domain of $$\sqrt {g(x)}$$ → square root of a quantity is defined when the quantity is non- negative, so solve $$g(x) \geq 0$$ to obtain the domain.

The domain of $$f(x) = \sqrt {7x - 1}$$ is

A $$x \epsilon \space (-\infty, 1/9]$$

B $$x \epsilon \space [1, \infty)$$

C $$x \epsilon \space [1/7, \infty)$$

D $$x \epsilon \space (-\infty, 8]$$

×

Since $$\sqrt {g(x)}$$ is defined when $$g(x) \geq 0,$$

$$7x - 1 \geq 0$$

Add 1 to both sides of inequality

$$7x - 1 + 1 \geq 1$$

$$7x \geq 1$$

Now divide by 7

$$x \geq \dfrac{1}{7}$$

The domain of $$f(x) = \sqrt {7x - 1}$$ is

A

$$x \epsilon \space (-\infty, 1/9]$$

.

B

$$x \epsilon \space [1, \infty)$$

C

$$x \epsilon \space [1/7, \infty)$$

D

$$x \epsilon \space (-\infty, 8]$$

Option C is Correct

Domain of Expression of the Form f(x)/g(x)

• To find the domain of the expression of the form $$\dfrac{f(x)}{g(x)}$$, we make sure that the denominator g(x) is not 0, otherwise the expression will not be defined.
• So domain of $$\dfrac{f(x)}{g(x)}$$ will be all real values of x except the roots of g(x) = 0.
• So solve for g(x) = 0 and the remaining real values will be the domain.
• We represent this as $$\{{x/g(x) \neq 0}\}$$

Find the domain of $$h(x) = \dfrac {3x^3+2x^2-x+5}{2x-3}$$

A  $$\left \{ x\mid x\,\neq \dfrac {2}{3} \right \}$$

B $$\left \{ x\mid x\,\neq \dfrac {3}{2} \right \}$$

C $$\left \{ x\mid x\,\neq 5 \right \}$$

D $$\left \{ x\mid x\,\neq 2 \right \}$$

×

Domain of expression of the form

$$\dfrac {f(x)}{g(x)}$$is $$\left \{ x\mid x\,\neq \text {root of }g(x) \right \}$$

In this case $$h(x)=\dfrac {3x^3+2x^2-x+5}{2x-3}=\dfrac {f(x)}{g(x)}$$

$$g(x)=0$$

$$\Rightarrow 2x-3=0$$

$$\Rightarrow x=\dfrac {3}{2}$$

$$\therefore$$ The domain of $$h(x)$$ is $$\left \{ x\mid x\,\neq \dfrac {3}{2} \right \}$$

Find the domain of $$h(x) = \dfrac {3x^3+2x^2-x+5}{2x-3}$$

A

$$\left \{ x\mid x\,\neq \dfrac {2}{3} \right \}$$

.

B

$$\left \{ x\mid x\,\neq \dfrac {3}{2} \right \}$$

C

$$\left \{ x\mid x\,\neq 5 \right \}$$

D

$$\left \{ x\mid x\,\neq 2 \right \}$$

Option B is Correct