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.

🔑Concepts

•

The Identity Function is defined by f:R→Rf: \mathbb{R} \to \mathbb{R} such that f(x)=xf(x) = x for each x∈Rx \in \mathbb{R}. The domain and range are both R\mathbb{R}, and the graph is a straight line passing through the origin at a 45∘45^\circ angle.

Graph of the Identity function f(x) = x showing a diagonal line through the origin.
•

The Modulus Function (or absolute value function) f(x)=∣x∣f(x) = |x| maps every real number to its non-negative value. The domain is R\mathbb{R} and the range is [0,∞)[0, \infty). The graph is V-shaped.

V-shaped graph of the modulus function.
•

The Signum Function f(x)=sgn(x)f(x) = \text{sgn}(x) outputs 11 for x>0x > 0, −1-1 for x<0x < 0, and 00 for x=0x = 0. It is used to extract the sign of a real number.

Step-like graph of the signum function.
•

The Greatest Integer Function f(x)=[x]f(x) = [x] (Floor function) returns the greatest integer less than or equal to xx. Its graph consists of horizontal line segments resembling steps.

Step graph of the Greatest Integer Function.

📐Formulae

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

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

Modulus Function: f(x)=∣x∣={x,x≥0−x,x<0f(x) = |x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}

Signum Function: f(x)=sgn(x)={∣x∣x,x≠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)≠0f(x) = \frac{p(x)}{q(x)}, q(x) \neq 0

💡Examples

Problem 1:

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

Solution:

  1. For any real number xx, the expression x−2x - 2 is always defined. Therefore, the Domain of ff is R\mathbb{R}.
  2. By definition of the modulus function, ∣x−2∣≥0|x - 2| \ge 0 for all x∈Rx \in \mathbb{R}.
  3. The smallest value occurs when x=2x = 2, where f(2)=∣2−2∣=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]+[−3.1]+sgn(−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: [−3.1][-3.1] is the greatest integer ≤−3.1\le -3.1, which is −4-4.
  3. Using the Signum Function definition: since −5<0-5 < 0, sgn(−5)=−1\text{sgn}(-5) = -1.
  4. Substituting these values into the expression: E=2+(−4)+(−1)=2−4−1=−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.

Problem 3:

Sketch the graph of the polynomial function f(x)=x2f(x) = x^2 for x∈Rx \in \mathbb{R}. State its domain and range.

Parabolic graph of f(x) = x^2.

Solution:

The domain is x∈Rx \in \mathbb{R} because the square of any real number is defined. Since the square of a number is always non-negative, the range is [0,∞)[0, \infty). The table of values includes (0,0),(1,1),(−1,1),(2,4),(−2,4)(0,0), (1,1), (-1,1), (2,4), (-2,4). Connecting these points gives a parabola opening upwards.

Explanation:

Polynomial functions of the form xnx^n behave differently based on whether nn is even or odd. For n=2n=2, the function is symmetric about the y-axis.

Problem 4:

Identify the features of the reciprocal function f(x)=1x,x≠0f(x) = \frac{1}{x}, x \neq 0 and visualize its graph.

Hyperbolic graph of the reciprocal function f(x)=1/x.

Solution:

The domain is R−{0}\mathbb{R} - \{0\} and the range is also R−{0}\mathbb{R} - \{0\}. As xx becomes very large, yy approaches 00. As xx approaches 00 from the positive side, yy approaches ∞\infty. The graph exists in the first and third quadrants.

Explanation:

This is a basic rational function where the denominator cannot be zero, creating a vertical asymptote at x=0x=0 and a horizontal asymptote at y=0y=0.