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Relations and Functions - Functions-advanced

Grade 9CBSE

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 special relation where every element x∈Ax \in A is associated with exactly one element y∈By \in B. Graphically, this is verified using the Vertical Line Test: if any vertical line intersects the graph more than once, the graph does not represent a function.

Graph of y = x^2 showing a vertical line intersecting at exactly one point, confirming it is a function.
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The Domain of a function is the set of all possible input values (xx) for which the function is defined. The Range is the set of all actual output values (yy). For a square root function like f(x)=xf(x) = \sqrt{x}, the domain is x≥0x \geq 0.

Graph of y = sqrt(x) starting from the origin and extending into the first quadrant.
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A Polynomial Function of degree 2 is called a Quadratic Function, expressed as f(x)=ax2+bx+cf(x) = ax^2 + bx + c. Its graph is a parabola. If a>0a > 0, the parabola opens upwards.

A parabola opening upwards representing a quadratic function.
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The Modulus Function (Absolute Value Function) is defined as f(x)=∣x∣f(x) = |x|, which returns xx if x≥0x \geq 0 and −x-x if x<0x < 0. The graph forms a 'V' shape with the vertex at the origin.

V-shaped graph of the modulus function f(x) = |x|.

📐Formulae

f:A→Bf: A \rightarrow B

y=f(x)y = f(x)

Identity Function: f(x)=x\text{Identity Function: } f(x) = x

Constant Function: f(x)=c\text{Constant Function: } f(x) = c

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

💡Examples

Problem 1:

If f(x)=3x2−5x+2f(x) = 3x^2 - 5x + 2, calculate the value of f(2)+f(−1)f(2) + f(-1).

Solution:

First, find f(2)f(2): f(2)=3(2)2−5(2)+2f(2) = 3(2)^2 - 5(2) + 2 f(2)=3(4)−10+2f(2) = 3(4) - 10 + 2 f(2)=12−10+2=4f(2) = 12 - 10 + 2 = 4

Next, find f(−1)f(-1): f(−1)=3(−1)2−5(−1)+2f(-1) = 3(-1)^2 - 5(-1) + 2 f(−1)=3(1)+5+2f(-1) = 3(1) + 5 + 2 f(−1)=3+5+2=10f(-1) = 3 + 5 + 2 = 10

Now, add the results: f(2)+f(−1)=4+10=14f(2) + f(-1) = 4 + 10 = 14

Explanation:

We evaluate the function by substituting the specific values of xx into the polynomial expression and then perform the required addition.

Problem 2:

Given f(x+2)=x2+5x−1f(x+2) = x^2 + 5x - 1, find the expression for f(x)f(x).

Solution:

Let t=x+2t = x + 2. This implies that x=t−2x = t - 2. Substitute x=t−2x = t - 2 into the expression for f(x+2)f(x+2): f(t)=(t−2)2+5(t−2)−1f(t) = (t-2)^2 + 5(t-2) - 1 Expand the terms: f(t)=(t2−4t+4)+(5t−10)−1f(t) = (t^2 - 4t + 4) + (5t - 10) - 1 Combine like terms: f(t)=t2+(−4t+5t)+(4−10−1)f(t) = t^2 + (-4t + 5t) + (4 - 10 - 1) f(t)=t2+t−7f(t) = t^2 + t - 7 Replacing tt back with xx: f(x)=x2+x−7f(x) = x^2 + x - 7

Explanation:

To find f(x)f(x) when f(expression)f(\text{expression}) is given, use a substitution variable to solve for the original variable in terms of the new one.

Problem 3:

Determine if the relation R={(1,5),(2,10),(3,15),(1,20)}R = \{(1, 5), (2, 10), (3, 15), (1, 20)\} is a function.

Solution:

Check the inputs (first elements of the ordered pairs): The inputs are 1,2,3,1, 2, 3, and 11. The input 11 is associated with two different outputs: 55 and 2020.

Explanation:

By definition, for a relation to be a function, each input in the domain must have exactly one unique output. Since the input 11 has two outputs, RR is not a function.

Problem 4:

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

Graph of the semi-circle y = sqrt(9-x^2) from x = -3 to 3.

Solution:

For the square root to be defined, 9−x2≥09 - x^2 \geq 0. This implies x2≤9x^2 \leq 9, so the Domain is [−3,3][-3, 3]. Since x2≥0x^2 \geq 0, the maximum value of f(x)f(x) is 9=3\sqrt{9} = 3 and the minimum is 0=0\sqrt{0} = 0. Thus, the Range is [0,3][0, 3].

Explanation:

The expression represents a semi-circle with radius 3 centered at the origin, lying above the x-axis.

Problem 5:

Determine the domain and range of the function f(x)=∣x∣−2f(x) = |x| - 2. Sketch the graph and find the value of xx for which f(x)=0f(x) = 0.

Graph of the absolute value function f(x) = |x| - 2 showing a V-shape shifted down by 2 units.

Solution:

  1. For the domain, since ∣x∣|x| is defined for all real numbers, the domain is x∈Rx \in \mathbb{R}.
  2. For the range, we know ∣x∣≥0|x| \geq 0 for all xx. Therefore, ∣x∣−2≥−2|x| - 2 \geq -2. The range is [−2,∞)[-2, \infty).
  3. To find xx when f(x)=0f(x) = 0, we set ∣x∣−2=0⇒∣x∣=2|x| - 2 = 0 \Rightarrow |x| = 2. This gives x=2x = 2 or x=−2x = -2.

Explanation:

The modulus function f(x)=∣x∣f(x) = |x| creates a V-shape with its vertex at the origin. Subtracting 2 from the function value shifts the entire graph downward by 2 units on the y-axis, making the new vertex (0,−2)(0, -2).