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Relations and Functions - Algebra of real functions

Grade 11CBSE

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

🔑Concepts

Definition of Real Functions: A function f:ABf: A \rightarrow B is called a real-valued function if BB is a subset of the set of real numbers R\mathbb{R}. If AA is also a subset of R\mathbb{R}, it is called a real function. Visually, the graph of such a function exists on a 2D coordinate plane where the horizontal axis represents the domain and the vertical axis represents the range.

Addition of Real Functions: For two real functions f:XRf: X \rightarrow \mathbb{R} and g:XRg: X \rightarrow \mathbb{R}, their sum (f+g):XR(f+g): X \rightarrow \mathbb{R} is defined by (f+g)(x)=f(x)+g(x)(f+g)(x) = f(x) + g(x). Graphically, this corresponds to 'pointwise addition' where at any given xx, the yy-coordinate of the new graph is the sum of the yy-coordinates of ff and gg.

Subtraction of Real Functions: The difference (fg):XR(f-g): X \rightarrow \mathbb{R} is defined by (fg)(x)=f(x)g(x)(f-g)(x) = f(x) - g(x). If we visualize the graphs of ff and gg, the resulting graph of fgf-g represents the vertical distance between the two curves at each xx. If the curves intersect, the value of the difference function is zero, crossing the xx-axis.

Multiplication by a Scalar: For a real function ff and a scalar α\alpha, the function (αf)(x)=αf(x)(\alpha f)(x) = \alpha \cdot f(x). Visually, if α>1\alpha > 1, the graph undergoes a vertical stretch; if 0<α<10 < \alpha < 1, it undergoes a vertical compression. If α\alpha is negative, the graph is reflected across the xx-axis.

Multiplication of Real Functions: The pointwise product (fg):XR(fg): X \rightarrow \mathbb{R} is defined by (fg)(x)=f(x)g(x)(fg)(x) = f(x)g(x). An important visual property is that if either f(x)=0f(x)=0 or g(x)=0g(x)=0 at a point xx, the product graph will have an xx-intercept (root) at that point.

Quotient of Real Functions: The quotient of two functions is defined as (fg)(x)=f(x)g(x)(\frac{f}{g})(x) = \frac{f(x)}{g(x)}, provided g(x)0g(x) \neq 0. On a graph, points where g(x)=0g(x) = 0 but f(x)0f(x) \neq 0 typically result in vertical asymptotes, where the curve approaches infinity as it nears that xx-value.

Domain Intersection Rule: When performing algebraic operations on two functions ff and gg, the domain of the resulting function (sum, difference, or product) is the intersection of their individual domains: DfDgD_f \cap D_g. For the quotient f/gf/g, we must further exclude all points where the denominator g(x)g(x) equals zero.

📐Formulae

Addition: (f+g)(x)=f(x)+g(x)(f+g)(x) = f(x) + g(x)

Subtraction: (fg)(x)=f(x)g(x)(f-g)(x) = f(x) - g(x)

Scalar Multiplication: (αf)(x)=αf(x)(\alpha f)(x) = \alpha f(x) for αR\alpha \in \mathbb{R}

Multiplication: (fg)(x)=f(x)g(x)(fg)(x) = f(x)g(x)

Quotient: (fg)(x)=f(x)g(x),g(x)0(\frac{f}{g})(x) = \frac{f(x)}{g(x)}, g(x) \neq 0

Domain of f±gf \pm g and fgfg: Df±g=Dfg=DfDgD_{f \pm g} = D_{fg} = D_f \cap D_g

Domain of f/gf/g: Df/g={xDfDg:g(x)0}D_{f/g} = \{x \in D_f \cap D_g : g(x) \neq 0\}

💡Examples

Problem 1:

Let f(x)=x2f(x) = x^2 and g(x)=2x+1g(x) = 2x + 1 be two real functions. Find (f+g)(x)(f + g)(x), (fg)(x)(f - g)(x), (fg)(x)(fg)(x), and (fg)(x)(\frac{f}{g})(x).

Solution:

  1. Addition: (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 \ 2. Subtraction: (fg)(x)=f(x)g(x)=x2(2x+1)=x22x1(f-g)(x) = f(x) - g(x) = x^2 - (2x + 1) = x^2 - 2x - 1 \ 3. Multiplication: (fg)(x)=f(x)g(x)=x2(2x+1)=2x3+x2(fg)(x) = f(x) \cdot g(x) = x^2(2x + 1) = 2x^3 + x^2 \ 4. Quotient: (fg)(x)=f(x)g(x)=x22x+1(\frac{f}{g})(x) = \frac{f(x)}{g(x)} = \frac{x^2}{2x + 1}, where 2x+10x122x + 1 \neq 0 \Rightarrow x \neq -\frac{1}{2}.

Explanation:

To solve these, we apply the pointwise algebraic definitions. For the quotient, we must identify the restriction on the domain where the denominator g(x)g(x) becomes zero.

Problem 2:

Given f(x)=xf(x) = \sqrt{x} and g(x)=xg(x) = x, find the domain of (fg)(x)(\frac{f}{g})(x).

Solution:

  1. Find individual domains: Df=[0,)D_f = [0, \infty) because the square root is defined for non-negative numbers. Dg=RD_g = \mathbb{R} because it is a linear polynomial. \ 2. Find the intersection: DfDg=[0,)D_f \cap D_g = [0, \infty). \ 3. Identify where g(x)=0g(x) = 0: g(x)=x=0g(x) = x = 0 at x=0x = 0. \ 4. Apply the quotient domain rule: Df/g=(DfDg){x:g(x)=0}=[0,){0}=(0,)D_{f/g} = (D_f \cap D_g) - \{x : g(x) = 0\} = [0, \infty) - \{0\} = (0, \infty).

Explanation:

The domain of a quotient function is the intersection of the domains of the numerator and denominator, excluding any points that make the denominator zero. Here, x=0x=0 is excluded even though it is in the domain of f(x)f(x).