Chapter 2.8 - Solved Examples on Functions

In the previous section, we saw algebra of real functions. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 2.34
Let R be the set of real numbers. Define the real function
f: R→R by f(x) = x + 10 and sketch the graph of this function.
Solution:
1. Given that f: R→R
So the input x values should be from R. The output f(x) values must also be present in R.
2. Let us input some convenient x values:
    ♦ When x = -10, f(x) = f(-10) = (-10 + 10) = 0
    ♦ When x = -9, f(x) = f(-9) = (-9 + 10) = 1
    ♦ When x = -7, f(x) = f(-7) = (-7 + 10) = 3
    ♦ When x = 0, f(x) = f(0) = (0 + 10) = 10
    ♦ When x = 2, f(x) = f(2) = (2 + 10) = 12
    ♦ When x = 3, f(x) = f(3) = (3 + 10) = 13
    ♦ When x = 4, f(x) = f(4) = (4 + 10) = 14
3. We can make a table using the above values:

Table.2.10

Such a table is convenient to plot the graph of the function.
4. The graph is shown in fig.2.20 below:

Fig.2.20

 

• We see that, the graph is a line.
• In our earlier analytical geometry classes, we have seen that:
Equation of a line takes the form y = mx + c, where m and c are constants.
• In our present case, the function is f(x) = x + 10. This is comparable to y = mx + c because:
    ♦ y is equivalent to f(x)
    ♦ m is equivalent to 1
    ♦ c is equivalent to 10
5. The function f defined by f(x) = mx + c, x ∈ R is called linear function.
• So our present case f(x) = x + 10 is a linear function.

Solved example 2.35
Let R be a relation from Q to Q defined by R = {(a, b) : a, b ∈ Q and a – b ∈ Z}. Show that
(i) (a, a) ∈ R for all a ∈ Q
(ii) (a, b) ∈ R implies that (b, a) ∈ R
(iii) (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R
Solution:
1. Q is the set of rational numbers. A relation R is defined from Q to Q.
• Details about this relation R can be written as follows:
   ♦ The set R contains all ordered pairs (a, b) such that,
   ♦ a is a rational number,
   ♦ b is also a rational number,
   ♦ a-b is an integer.
2. Recall that,
   ♦ rational numbers can have a decimal part
   ♦ integers do not have a decimal part
3. We can write infinite pairs of rational numbers in such a way that their differences are integers.
• Let us see some examples:
   ♦ For the pair (3.157, 11.157), difference = (3.157 – 11.157) = -8
   ♦ For the pair (-2.73, 10.27), difference = (-2.73 – 10.27) = 13
• All such pairs are eligible to be included in the set R
• Now we can write the answers of the given questions.
Part (i):
• The pairs are in the form (a, a) and a is a rational number.
   ♦ Here both the elements are the same. Their difference (a-a) will be zero.
• Zero is an integer. So all pairs in the form (a, a), where a is a rational number, are eligible to be included in R.
• We can write: (a, a) ∈ R for all a ∈ Q
Part (ii):
• If a and b are such that a - b is an integer, then b – a will also be an integer.
An example:
• Consider the pair (3.157, 11.157)
   ♦ The difference (3.157 – 11.157) = -8
   ♦ The difference (11.157 – 3.157) = 8
Another example:
• Consider the pair (-2.73, 10.27)
   ♦ The difference (-2.73 – 10.27) = -13
   ♦ The difference (10.27 –  - 2.73) = 13
• So (b, a) will be eligible to be included in R
• We can write: (a,b) ∈ R implies that (b, a) ∈ R
Part (iii):
• We have to prove that, (a - c) is an integer if
   ♦ (a – b) gives an integer
   ♦ (b – c) also gives an integer,
• Let us see an example:
   ♦ Consider the pair (-2.73, 10.27)
   ♦ The difference (a – b) = (-2.73 – 10.27) = -13
   ♦ Consider the pair (10.27, 6.27)
   ♦ The difference (b – c) = (10.27 – 6.27) = 4
   ♦ Now the difference (a – c) = (-2.73 – 6.27) = -9
• So (a,c) is eligible to be included in R.
• We can prove this algebraically also:
   ♦ (a - b) + (b - c) = a - b + b - c = (a - c)
   ♦ (-2.73 – 10.27) + (10.27 – 6.27) = (-2.73 – 6.27)
• We can write: (a, b) ∈ R and (b, c) ∈ R implies that (a, c) ∈ R

Solved example 2.36
Let f = {(1,1), (2,3), (0, –1), (–1, –3)} be a linear function from Z into Z. Find f(x).
Solution:
1. Given that f is a linear function. So it will be in the form: f(x) = mx + c
All ordered pairs in the set f will satisfy this relation.
2. Consider the pair (1, 1). We get: 1 = m × 1 + c
⇒ 1 = m + c   
3. Consider the pair (2, 3). We get: 3 = m × 2 + c
⇒ 3 = 2m + c
4. Subtracting (2) from (3), we get:
3 - 1 = (2m + c) - (m + c)
⇒ 2 = m
5. Substituting for m in (2), we get: 1 = 2 + c
⇒ c = -1
6. Substituting for m and c in (1), we get: f(x) = 2x - 1
7. Note: We used the ordered pairs (1, 1) and (2, 3)
   ♦ It would be much more easier if we use (1, 1) and (0, -1)
   ♦ This is because, the term with m will become zero.

Solved example 2.37
Find the domain of the function $\mathbf\small{\rm{f(x)=\frac{x^2+3x+5}{x^2-5x+4}}}$
Solution:
1. Consider the denominator x² - 5x + 4
• If this denominator become zero, f(x) cannot be defined. So we must not input those x values which make x² - 5x + 4 zero.
2. So we need to find those x values which will make x² - 5x + 4 zero.
• For that, we equate it to zero and solve for x.
x² - 5x + 4 = 0
⇒ x² - 5x = -4
⇒ x² - 5x + (⁵⁄₂)² = -4 + (⁵⁄₂)²
⇒ (x - ⁵⁄₂)² = -4 + ²⁵⁄₄
⇒ (x - ⁵⁄₂)² = ⁹⁄₄
⇒ x- ⁵⁄₂ = ± ³⁄₂
⇒ x = (⁵⁄₂ + ³⁄₂) or (⁵⁄₂ - ³⁄₂)
⇒ x = ⁸⁄₂ or ²⁄₂
⇒ x = 4 or 1
3. So it is clear that, we must not use 4 and 1 as input x.
• We can write: domain = R - {4, 1}
• That means, domain is the set obtained by performing the 'difference' operation on the sets R and {4, 1}
   ♦ See difference of two sets.

Solved example 2.38
The function f is defined by:
$f(x) =
\begin{cases}
1-x,  & \text{if}\; x < 0 \\
1,  & \text{if}\; x = 0 \\
x+1, & \text{if} \; x > 0
\end{cases}$
Draw the graph of f(x).
Solution:
1. If we input an x value which is less than 0, that x value will be processed according to the function: f(x) = 1 - x
• Let us try some x values which are less than 0:
   ♦ When x = -5, f(x) = 1 - x = 1 - (-5) = 6
   ♦ When x = -3, f(x) = 1 - x = 1 - (-3) = 4
2. If we input an x value which is equal to 0, that x value will be processed according to the function: f(x) = 1
• Here there is no need to try different x values. Only one input x is possible and that is '0'.
• Also it is given that, when x = 0, f(x) = a constant, which is '1'.
• So there is no need to see how the x value is processed.
3. If we input an x value which is greater than 0, that x value will be processed according to the function: f(x) = x + 1
• Let us try some x values which are less than 0:
   ♦ When x = 2, f(x) = x + 1 = 2 + 1 = 3
   ♦ When x = 7, f(x) = x + 1 = 7 + 1 = 8
4. Based on the above steps, we can make a table:

Table 2.11

5. Using the table, we can draw the graph:

Fig.2.21

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Link to some more solved examples is given below:

Solved examples 2.39 to 2.49

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We have completed a discussion on relations and functions. In the next chapter, we will see trigonometric functions.

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