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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

     

Illustration Questions

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 \(ax + b \geq 0\).

Illustration Questions

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

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.

Illustration Questions

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

Illustration Questions

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 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 \(\{{x/g(x) \neq 0}\}\)

Illustration Questions

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

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