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Sets and Functions - Functions, their Domain, Range, and Graphs

Grade 11ICSE

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

🔑Concepts

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A function f:A→Bf: A \rightarrow B is a rule that assigns each element xx in set AA (Domain) to exactly one element yy in set BB (Codomain). The actual set of outputs produced is the Range. Visually, the vertical line test determines if a graph represents a function: a vertical line must cross the graph at most once.

Graph of y = x^2 passing the vertical line test at x = 2.
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The Constant Function is defined by f(x)=cf(x) = c for all x∈Rx \in \mathbb{R}, where cc is a constant. Its graph is a horizontal line parallel to the x-axis. The domain is R\mathbb{R} and the range is the singleton set {c}\{c\}.

Graph of a constant function y = 3 as a horizontal line.
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The Identity Function f(x)=xf(x) = x assigns every real number to itself. Its graph is a straight line passing through the origin at an angle of 45∘45^\circ with the positive x-axis. Domain = R\mathbb{R}, Range = R\mathbb{R}.

Graph of the identity function f(x) = x.
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The Modulus (Absolute Value) Function f(x)=∣x∣f(x) = |x| returns the non-negative value of xx. The graph is V-shaped with the vertex at the origin. Domain is R\mathbb{R}, and Range is [0,∞)[0, \infty).

V-shaped graph of the modulus function y = |x|.

📐Formulae

f:A→Bf: A \rightarrow B where ∀x∈A,∃!y∈B\forall x \in A, \exists! y \in B

Domain of P(x)Q(x)={x∈R:Q(x)≠0}\text{Domain of } \frac{P(x)}{Q(x)} = \{x \in \mathbb{R} : Q(x) \neq 0\}

Domain of f(x)={x∈R:f(x)≥0}\text{Domain of } \sqrt{f(x)} = \{x \in \mathbb{R} : f(x) \geq 0\}

∣x∣=x2={xif x≥0−xif x<0|x| = \sqrt{x^2} = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

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

[x]=n  ⟺  n≤x<n+1,n∈Z[x] = n \iff n \leq x < n+1, n \in \mathbb{Z}

(fg)(x)=f(x)g(x), provided g(x)≠0(\frac{f}{g})(x) = \frac{f(x)}{g(x)}, \text{ provided } g(x) \neq 0

💡Examples

Problem 1:

Find the domain and range of the function f(x)=9−x2f(x) = \sqrt{9 - x^2}.

Solution:

  1. For the function to be defined, the expression inside the square root must be non-negative: 9−x2≥09 - x^2 \geq 0.
  2. Solve the inequality: x2≤9x^2 \leq 9, which gives −3≤x≤3-3 \leq x \leq 3. Thus, Domain =[−3,3]= [-3, 3].
  3. To find the range, let y=9−x2y = \sqrt{9 - x^2}. Since it is a square root, y≥0y \geq 0.
  4. Squaring both sides: y2=9−x2  ⟹  x2=9−y2y^2 = 9 - x^2 \implies x^2 = 9 - y^2.
  5. Since x2≥0x^2 \geq 0, we have 9−y2≥0  ⟹  y2≤99 - y^2 \geq 0 \implies y^2 \leq 9, so −3≤y≤3-3 \leq y \leq 3.
  6. Combining y≥0y \geq 0 and −3≤y≤3-3 \leq y \leq 3, the Range is [0,3][0, 3].

Explanation:

The domain is restricted by the square root condition (radicand ≥0\geq 0). The range is restricted by both the output of the square root (always non-negative) and the maximum value of the radicand.

Problem 2:

Find the domain of f(x)=x2+3x+5x2−5x+4f(x) = \frac{x^2 + 3x + 5}{x^2 - 5x + 4}.

Solution:

  1. The function is a rational function, so it is defined for all xx except where the denominator equals zero.
  2. Set the denominator to zero: x2−5x+4=0x^2 - 5x + 4 = 0.
  3. Factor the quadratic: (x−4)(x−1)=0(x - 4)(x - 1) = 0.
  4. Find the roots: x=4x = 4 and x=1x = 1.
  5. Therefore, the domain is the set of all real numbers except 11 and 44.
  6. In interval notation: Domain =R−{1,4}= \mathbb{R} - \{1, 4\} or (−∞,1)∪(1,4)∪(4,∞)(-\infty, 1) \cup (1, 4) \cup (4, \infty).

Explanation:

For rational functions, the numerator can be anything, but the denominator cannot be zero as division by zero is undefined in real numbers.

Problem 3:

Draw the graph and determine the range of the function f(x)=2x−3f(x) = 2x - 3 for the domain D={x∈R:0≤x≤4}D = \{x \in \mathbb{R} : 0 \leq x \leq 4\}.

Graph of f(x) = 2x - 3 showing a line segment from (0, -3) to (4, 5).

Solution:

  1. Find the values at the boundaries: When x=0,f(0)=2(0)−3=−3x = 0, f(0) = 2(0) - 3 = -3. When x=4,f(4)=2(4)−3=5x = 4, f(4) = 2(4) - 3 = 5.
  2. Since it is a linear function, the graph is a line segment connecting (0,−3)(0, -3) and (4,5)(4, 5).
  3. Range: Looking at the y-values, the range is [−3,5][-3, 5].

Explanation:

For a linear function f(x)=ax+bf(x) = ax + b over a closed interval [x1,x2][x_1, x_2], the range is the interval between f(x1)f(x_1) and f(x2)f(x_2).

Problem 4:

Identify the domain and draw the graph of the reciprocal function f(x)=1xf(x) = \frac{1}{x}.

Graph of the reciprocal function y = 1/x showing two hyperbola branches.

Solution:

  1. The function is defined for all real numbers except where the denominator is zero. Thus, x≠0x \neq 0.
  2. Domain: R−{0}\mathbb{R} - \{0\}.
  3. As x→∞,f(x)→0x \rightarrow \infty, f(x) \rightarrow 0. As x→0+,f(x)→∞x \rightarrow 0^+, f(x) \rightarrow \infty. The graph consists of two branches in the first and third quadrants.

Explanation:

The graph is a rectangular hyperbola with asymptotes at x=0x=0 and y=0y=0. It never touches either axis.