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Relations and Functions - Functions

Grade 11CBSE

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

🔑Concepts

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A function ff from a set AA to a set BB is a specific type of relation where every element of set AA has exactly one image in set BB. In a mapping diagram, this means every element in the domain must have exactly one arrow originating from it.

Mapping diagram showing a function from set A to set B where each element has exactly one image.
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The Square Function f(x)=x2f(x) = x^2 maps every real number to its square. Its domain is RR and its range is [0,∞)[0, \infty). The graph is a parabola opening upwards with the vertex at the origin.

Graph of the square function f(x) = x^2 showing a parabola.
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The Modulus Function f(x)=∣x∣f(x) = |x| is defined as xx if x≥0x \geq 0 and −x-x if x<0x < 0. Geometrically, it represents a 'V' shaped graph symmetric about the y-axis.

Graph of the modulus function f(x) = |x|.
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A Rational Function is defined as f(x)=p(x)q(x)f(x) = \frac{p(x)}{q(x)}, where p(x)p(x) and q(x)q(x) are polynomial functions and q(x)≠0q(x) \neq 0. The domain excludes values of xx that make the denominator zero.

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

📐Formulae

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

Constant Function: f(x)=cf(x) = c, where cc is a constant

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

Signum Function: f(x)={1,x>00,x=0−1,x<0=∣x∣x,x≠0f(x) = \begin{cases} 1, & x > 0 \\ 0, & x = 0 \\ -1, & x < 0 \end{cases} = \frac{|x|}{x}, x \neq 0

Greatest Integer Function: f(x)=[x]f(x) = [x], where n≤x<n+1⇒[x]=nn \leq x < n+1 \Rightarrow [x] = n

Domain of fg\frac{f}{g}: Df∩Dg−{x:g(x)=0}D_f \cap D_g - \{x : g(x) = 0\}

Domain of f(x)\sqrt{f(x)}: {x∈Df:f(x)≥0}\{x \in D_f : f(x) \geq 0\}

💡Examples

Problem 1:

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

Solution:

  1. For f(x)f(x) to be defined as a real function, the expression inside the square root must be non-negative: 9−x2≥09 - x^2 \geq 0.
  2. Factoring the inequality: (3−x)(3+x)≥0(3 - x)(3 + x) \geq 0. This implies −3≤x≤3-3 \leq x \leq 3. So, Domain =[−3,3]= [-3, 3].
  3. To find the range, let y=9−x2y = \sqrt{9 - x^2}. Since square roots are non-negative, y≥0y \geq 0.
  4. Squaring both sides: y2=9−x2⇒x2=9−y2y^2 = 9 - x^2 \Rightarrow x^2 = 9 - y^2.
  5. Since x2≥0x^2 \geq 0, then 9−y2≥0⇒y2≤9⇒−3≤y≤39 - y^2 \geq 0 \Rightarrow y^2 \leq 9 \Rightarrow -3 \leq y \leq 3.
  6. Combining y≥0y \geq 0 and −3≤y≤3-3 \leq y \leq 3, we get 0≤y≤30 \leq y \leq 3. So, Range =[0,3]= [0, 3].

Explanation:

We determine the domain by ensuring the radicand of the square root is non-negative. For the range, we solve for xx in terms of yy and apply the constraints of the square root's output.

Problem 2:

Let f(x)=x+1f(x) = x + 1 and g(x)=2x−3g(x) = 2x - 3. Find (f+g)(x)(f + g)(x), (f−g)(x)(f - g)(x), and (fg)(x)(\frac{f}{g})(x).

Solution:

  1. Addition: (f+g)(x)=f(x)+g(x)=(x+1)+(2x−3)=3x−2(f + g)(x) = f(x) + g(x) = (x + 1) + (2x - 3) = 3x - 2.
  2. Subtraction: (f−g)(x)=f(x)−g(x)=(x+1)−(2x−3)=x+1−2x+3=−x+4(f - g)(x) = f(x) - g(x) = (x + 1) - (2x - 3) = x + 1 - 2x + 3 = -x + 4.
  3. Division: (fg)(x)=f(x)g(x)=x+12x−3(\frac{f}{g})(x) = \frac{f(x)}{g(x)} = \frac{x + 1}{2x - 3}.
  4. Condition for division: The denominator cannot be zero, so 2x−3≠0⇒x≠322x - 3 \neq 0 \Rightarrow x \neq \frac{3}{2}.

Explanation:

This demonstrates the algebraic operations on functions. For addition and subtraction, we combine like terms. For division, we must explicitly state the restriction on the domain where the divisor is zero.

Problem 3:

Identify the domain and range of the real function f(x)=−∣x∣f(x) = -|x|. Sketch its graph.

Graph of f(x) = -|x| showing an inverted V shape.

Solution:

  1. For f(x)=−∣x∣f(x) = -|x|, xx can be any real number. Hence, Domain =R= R.
  2. Since ∣x∣≥0|x| \geq 0 for all x∈Rx \in R, multiplying by −1-1 gives −∣x∣≤0-|x| \leq 0. Thus, the values of f(x)f(x) are always non-positive. Range =(−∞,0]= (-\infty, 0].
  3. The graph is the reflection of y=∣x∣y = |x| in the x-axis.

Explanation:

The modulus ∣x∣|x| is always non-negative. Applying a negative sign flips the V-shape downwards, restricting the output to negative values and zero.

Problem 4:

Find the domain of the function f(x)=x2+2x+1x2−8x+12f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}.

Graph of the rational function showing vertical asymptotes at x=2 and x=6.

Solution:

  1. The function is defined for all xx such that the denominator x2−8x+12≠0x^2 - 8x + 12 \neq 0.
  2. Factorize the denominator: x2−6x−2x+12=(x−6)(x−2)x^2 - 6x - 2x + 12 = (x-6)(x-2).
  3. Set (x−6)(x−2)=0⇒x=6,2(x-6)(x-2) = 0 \Rightarrow x = 6, 2.
  4. Therefore, the domain is R−{2,6}R - \{2, 6\}.

Explanation:

A rational function is undefined when its denominator is zero. By finding the roots of the quadratic denominator, we identify the values to exclude from the set of real numbers.