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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.

πŸ”‘Concepts

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A function ff from a set AA to a set BB is a special relation where every element x∈Ax \in A is associated with exactly one element y∈By \in B. Visually, this means in an arrow diagram, every element in the domain has exactly one outgoing arrow. In a coordinate plane, this satisfies the Vertical Line Test: no vertical line intersects the graph of a function at more than one point.

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The Domain of a function is the set of all possible input values (xx) for which the function is defined. For real-valued functions, we must ensure that denominators are non-zero (to avoid division by zero) and expressions under even roots (like x\sqrt{x}) are non-negative. On a graph, the domain is the horizontal span of the curve along the xx-axis.

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The Range is the set of all actual output values (f(x)f(x) or yy) that the function produces. The Codomain is the set BB that contains the range. Visually, the range is represented by the vertical span of the graph along the yy-axis.

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The Identity Function is defined by f(x)=xf(x) = x. Its graph is a straight line passing through the origin at a 45∘45^{\circ} angle to the xx-axis, covering all real numbers for both domain and range. The Constant Function f(x)=cf(x) = c results in a horizontal line parallel to the xx-axis, where the range is the singleton set {c}\{c\}.

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The Modulus Function f(x)=∣x∣f(x) = |x| returns the absolute value of xx. The graph forms a symmetric 'V' shape with the vertex at the origin (0,0)(0,0). Its domain is R\mathbb{R} and its range is [0,∞)[0, \infty). It is defined piecewise as xx if xβ‰₯0x \geq 0 and βˆ’x-x if x<0x < 0.

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The Signum Function f(x)=∣x∣xf(x) = \frac{|x|}{x} for xβ‰ 0x \neq 0 and f(0)=0f(0) = 0 provides the sign of the input. Its graph consists of two horizontal rays: y=1y = 1 for x>0x > 0 and y=βˆ’1y = -1 for x<0x < 0, along with an isolated point at (0,0)(0,0). The range is specifically the set {βˆ’1,0,1}\{-1, 0, 1\}.

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The Greatest Integer Function f(x)=[x]f(x) = [x] (also known as the floor function) outputs the greatest integer less than or equal to xx. The graph resembles a series of 'steps' or a 'staircase,' where each step is closed on the left end and open on the right end (e.g., the interval [1,2)[1, 2) stays at y=1y=1).

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Algebra of Functions: For two functions ff and gg, we can define (f+g)(x)=f(x)+g(x)(f+g)(x) = f(x) + g(x), (fβˆ’g)(x)=f(x)βˆ’g(x)(f-g)(x) = f(x) - g(x), and (fβ‹…g)(x)=f(x)β‹…g(x)(f \cdot g)(x) = f(x) \cdot g(x). The domain of these new functions is the intersection of the domains of ff and gg, denoted as Df∩DgD_f \cap D_g.

πŸ“Formulae

f:Aβ†’Bf: A \rightarrow B where βˆ€x∈A,βˆƒ!y∈B\forall x \in A, \exists! y \in B

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

DomainΒ ofΒ f(x)={x∈R:f(x)β‰₯0}\text{Domain of } \sqrt{f(x)} = \{x \in \mathbb{R} : f(x) \geq 0\}

∣x∣=x2={xifΒ xβ‰₯0βˆ’xifΒ x<0|x| = \sqrt{x^2} = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

sgn(x)={1ifΒ x>00ifΒ x=0βˆ’1ifΒ 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]=nβ€…β€ŠβŸΊβ€…β€Šn≀x<n+1,n∈Z[x] = n \iff n \leq x < n+1, n \in \mathbb{Z}

(fg)(x)=f(x)g(x),Β providedΒ g(x)β‰ 0(\frac{f}{g})(x) = \frac{f(x)}{g(x)}, \text{ provided } g(x) \neq 0

πŸ’‘Examples

Problem 1:

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

Solution:

  1. For the function to be defined, the expression inside the square root must be non-negative: 9βˆ’x2β‰₯09 - x^2 \geq 0.
  2. Solve the inequality: x2≀9x^2 \leq 9, which gives βˆ’3≀x≀3-3 \leq x \leq 3. Thus, Domain =[βˆ’3,3]= [-3, 3].
  3. To find the range, let y=9βˆ’x2y = \sqrt{9 - x^2}. Since it is a square root, yβ‰₯0y \geq 0.
  4. Squaring both sides: y2=9βˆ’x2β€…β€ŠβŸΉβ€…β€Šx2=9βˆ’y2y^2 = 9 - x^2 \implies x^2 = 9 - y^2.
  5. Since x2β‰₯0x^2 \geq 0, we have 9βˆ’y2β‰₯0β€…β€ŠβŸΉβ€…β€Šy2≀99 - y^2 \geq 0 \implies y^2 \leq 9, so βˆ’3≀y≀3-3 \leq y \leq 3.
  6. Combining yβ‰₯0y \geq 0 and βˆ’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 β‰₯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+5x2βˆ’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: x2βˆ’5x+4=0x^2 - 5x + 4 = 0.
  3. Factor the quadratic: (xβˆ’4)(xβˆ’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 =Rβˆ’{1,4}= \mathbb{R} - \{1, 4\} or (βˆ’βˆž,1)βˆͺ(1,4)βˆͺ(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.