krit.club logo

Relations and Functions - Some functions and their graphs: Identity, Constant, Polynomial, Rational, Modulus, Signum, Greatest Integer Function

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

β€’

The Identity Function is defined as f:Rβ†’Rf: \mathbb{R} \to \mathbb{R} by y=f(x)=xy = f(x) = x for every x∈Rx \in \mathbb{R}. The domain and range are both the set of real numbers R\mathbb{R}. Visually, the graph is a straight line passing through the origin (0,0)(0, 0), making a 45∘45^\circ angle with both the x and y axes, extending infinitely in both directions.

β€’

The Constant Function is defined as f:R→Rf: \mathbb{R} \to \mathbb{R} by y=f(x)=cy = f(x) = c, where cc is a constant. The domain is R\mathbb{R} while the range is the singleton set {c}\{c\}. The graph is a horizontal line parallel to the x-axis, crossing the y-axis at the point (0,c)(0, c). If c>0c > 0, the line is above the x-axis; if c<0c < 0, it is below.

β€’

A Polynomial Function is defined by f(x)=a0+a1x+a2x2+β‹―+anxnf(x) = a_0 + a_1x + a_2x^2 + \dots + a_nx^n, where nn is a non-negative integer and ai∈Ra_i \in \mathbb{R}. Common examples include the squaring function f(x)=x2f(x) = x^2 (a parabola opening upwards with vertex at the origin) and the cubing function f(x)=x3f(x) = x^3 (an S-shaped curve passing through the origin and extending into the 1st and 3rd quadrants).

β€’

A Rational Function is of the form f(x)=g(x)h(x)f(x) = \frac{g(x)}{h(x)}, where g(x)g(x) and h(x)h(x) are polynomial functions and h(x)β‰ 0h(x) \neq 0. For the reciprocal function f(x)=1xf(x) = \frac{1}{x}, the domain and range are both Rβˆ’{0}\mathbb{R} - \{0\}. Its graph is a rectangular hyperbola that approaches the axes but never touches them, appearing in the first and third quadrants.

β€’

The Modulus Function f:Rβ†’Rf: \mathbb{R} \to \mathbb{R} is defined by f(x)=∣x∣f(x) = |x|, which equals xx if xβ‰₯0x \ge 0 and βˆ’x-x if x<0x < 0. The domain is R\mathbb{R} and the range is the set of non-negative real numbers [0,∞)[0, \infty). The graph is a V-shaped figure with the vertex at the origin, perfectly symmetric about the y-axis.

β€’

The Signum Function f:Rβ†’Rf: \mathbb{R} \to \mathbb{R} is defined as f(x)=1f(x) = 1 if x>0x > 0, f(x)=0f(x) = 0 if x=0x = 0, and f(x)=βˆ’1f(x) = -1 if x<0x < 0. The domain is R\mathbb{R} and the range is {βˆ’1,0,1}\{-1, 0, 1\}. The graph consists of two horizontal half-lines (rays) at y=1y = 1 and y=βˆ’1y = -1 starting from the y-axis (excluding the y-axis points) and a single point at the origin (0,0)(0, 0).

β€’

The Greatest Integer Function f:Rβ†’Rf: \mathbb{R} \to \mathbb{R} is defined by f(x)=[x]f(x) = [x], which denotes the greatest integer less than or equal to xx. For example, [2.3]=2[2.3] = 2 and [βˆ’2.3]=βˆ’3[-2.3] = -3. The domain is R\mathbb{R} and the range is the set of integers Z\mathbb{Z}. The graph looks like a staircase; each step is one unit long, having a closed circle on the left end and an open circle on the right end.

πŸ“Formulae

Identity Function: f(x)=xf(x) = x

Constant Function: f(x)=cf(x) = c

Modulus Function: f(x)=∣x∣={x,xβ‰₯0βˆ’x,x<0f(x) = |x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}

Signum Function: f(x)=sgn(x)={∣x∣x,xβ‰ 00,x=0f(x) = \text{sgn}(x) = \begin{cases} \frac{|x|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}

Greatest Integer Function: f(x)=[x]=nf(x) = [x] = n, where n≀x<n+1n \le x < n+1 and n∈Zn \in \mathbb{Z}

Rational Function condition: f(x)=p(x)q(x),q(x)β‰ 0f(x) = \frac{p(x)}{q(x)}, q(x) \neq 0

πŸ’‘Examples

Problem 1:

Find the domain and range of the function f(x)=∣xβˆ’2∣f(x) = |x - 2|.

Solution:

  1. For any real number xx, the expression xβˆ’2x - 2 is always defined. Therefore, the Domain of ff is R\mathbb{R}.
  2. By definition of the modulus function, ∣xβˆ’2∣β‰₯0|x - 2| \ge 0 for all x∈Rx \in \mathbb{R}.
  3. The smallest value occurs when x=2x = 2, where f(2)=∣2βˆ’2∣=0f(2) = |2 - 2| = 0. As xx increases or decreases from 2, f(x)f(x) increases towards ∞\infty.
  4. Thus, the Range is [0,∞)[0, \infty).

Explanation:

The modulus function always produces non-negative outputs, shifting the vertex of the V-shaped graph to (2,0)(2, 0).

Problem 2:

Evaluate the value of the expression E=[2.7]+[βˆ’3.1]+sgn(βˆ’5)E = [2.7] + [-3.1] + \text{sgn}(-5).

Solution:

  1. Using the definition of the Greatest Integer Function: [2.7][2.7] is the greatest integer ≀2.7\le 2.7, which is 22.
  2. For the negative value: [βˆ’3.1][-3.1] is the greatest integer β‰€βˆ’3.1\le -3.1, which is βˆ’4-4.
  3. Using the Signum Function definition: since βˆ’5<0-5 < 0, sgn(βˆ’5)=βˆ’1\text{sgn}(-5) = -1.
  4. Substituting these values into the expression: E=2+(βˆ’4)+(βˆ’1)=2βˆ’4βˆ’1=βˆ’3E = 2 + (-4) + (-1) = 2 - 4 - 1 = -3.

Explanation:

This problem applies the step-wise definition of the Greatest Integer Function and the piecewise definition of the Signum Function.

Some functions and their graphs: Identity, Constant, Polynomial, Rational, Modulus, Signum…