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Relations and Functions - Some Functions and their Graphs-advanced

Grade 9CBSE

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

🔑Concepts

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The Identity Function is defined by f(x)=xf(x) = x for all x∈Rx \in \mathbb{R}. Its graph is a straight line passing through the origin at an angle of 45∘45^\circ with the positive xx-axis. The domain and range are both the set of all real numbers R\mathbb{R}.

Graph of the identity function f(x) = x
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The Square Function is defined by f(x)=x2f(x) = x^2. It produces a symmetric curve called a parabola. Since any real number squared is non-negative, the range of this function is [0,∞)[0, \infty), while the domain is R\mathbb{R}.

Graph of the square function f(x) = x^2
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The Constant Function is defined by f(x)=cf(x) = c, where cc is a fixed real number. Regardless of the input xx, the output remains cc. Its graph is a horizontal line parallel to the xx-axis.

Graph of a constant function f(x) = 3
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The Cube Function is defined by f(x)=x3f(x) = x^3. It is an odd function because f(−x)=−f(x)f(-x) = -f(x), meaning the graph is symmetric with respect to the origin. The domain and range are both (−∞,∞)(-\infty, \infty).

Graph of the cube function f(x) = x^3

📐Formulae

f:A→B is a function if ∀x∈A,∃!y∈B s.t. (x,y)∈ff: A \rightarrow B \text{ is a function if } \forall x \in A, \exists ! y \in B \text{ s.t. } (x, y) \in f

Modulus Function: f(x)={xif x≥0−xif x<0\text{Modulus Function: } f(x) = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}

Signum Function: f(x)={1if x>00if x=0−1if x<0\text{Signum Function: } f(x) = \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -1 & \text{if } x < 0 \end{cases}

Linear Function: f(x)=mx+c\text{Linear Function: } f(x) = mx + c

Greatest Integer Function: [x]≤x<[x]+1\text{Greatest Integer Function: } [x] \le x < [x] + 1

💡Examples

Problem 1:

Find the domain and range of the real function f(x)=x−3f(x) = \sqrt{x - 3}.

Solution:

To find the domain, the expression under the square root must be non-negative: x−3≥0x - 3 \ge 0 x≥3x \ge 3 So, Domain =[3,∞)= [3, \infty). For the range, as xx takes values from 33 to ∞\infty, x−3\sqrt{x-3} takes all non-negative real values. Range =[0,∞)= [0, \infty).

Explanation:

Square root functions are only defined for non-negative values in the set of real numbers. The output of a principal square root is always non-negative.

Problem 2:

Sketch the graph and find the domain and range of f(x)=∣x+2∣f(x) = |x + 2|.

Solution:

The function can be rewritten as: f(x)={x+2if x+2≥0−(x+2)if x+2<0f(x) = \begin{cases} x + 2 & \text{if } x + 2 \ge 0 \\ -(x + 2) & \text{if } x + 2 < 0 \end{cases} Which simplifies to: f(x)={x+2if x≥−2−x−2if x<−2f(x) = \begin{cases} x + 2 & \text{if } x \ge -2 \\ -x - 2 & \text{if } x < -2 \end{cases} Domain: R\mathbb{R} (All real numbers). Range: [0,∞)[0, \infty) because the absolute value is never negative.

Explanation:

The graph is a V-shape shifted 22 units to the left on the xx-axis. The vertex is at (−2,0)(-2, 0).

Problem 3:

If f(x)=x2f(x) = x^2 and g(x)=2x+1g(x) = 2x + 1 are two real functions, find (f+g)(x)(f + g)(x) and (f⋅g)(x)(f \cdot g)(x).

Solution:

(f+g)(x)=f(x)+g(x)=x2+2x+1=(x+1)2(f + g)(x) = f(x) + g(x) = x^2 + 2x + 1 = (x + 1)^2 (f⋅g)(x)=f(x)⋅g(x)=x2(2x+1)=2x3+x2(f \cdot g)(x) = f(x) \cdot g(x) = x^2(2x + 1) = 2x^3 + x^2

Explanation:

Operations on functions are performed by adding or multiplying their respective algebraic expressions for the same input xx.

Problem 4:

Sketch the graph of the function f(x)=x+1f(x) = x + 1 and identify its intercepts.

Graph of f(x) = x + 1 showing intercepts at (0,1) and (-1,0)

Solution:

  1. This is a linear function of the form f(x)=mx+cf(x) = mx + c where m=1m=1 and c=1c=1.
  2. To find the yy-intercept, set x=0x = 0: f(0)=0+1=1f(0) = 0 + 1 = 1. Point is (0,1)(0, 1).
  3. To find the xx-intercept, set f(x)=0f(x) = 0: 0=x+1⇒x=−10 = x + 1 \Rightarrow x = -1. Point is (−1,0)(-1, 0).
  4. Drawing a line through these points gives the graph of the function.

Explanation:

Linear functions always result in straight lines. The constant term represents the shift along the yy-axis.

Problem 5:

Draw the graph of the absolute value function f(x)=∣x−1∣f(x) = |x - 1|.

Graph of f(x) = |x - 1| with vertex at (1,0)

Solution:

  1. The base function is ∣x∣|x|, which has a 'V' shape with a vertex at (0,0)(0,0).
  2. The function f(x)=∣x−1∣f(x) = |x - 1| shifts the vertex of the graph 1 unit to the right.
  3. When x=1x=1, f(1)=0f(1)=0. For x>1x > 1, f(x)=x−1f(x) = x-1. For x<1x < 1, f(x)=−(x−1)=1−xf(x) = -(x-1) = 1-x.
  4. The domain is R\mathbb{R} and the range is [0,∞)[0, \infty).

Explanation:

The expression ∣x−h∣|x-h| results in a horizontal translation of the modulus graph by hh units.