Question
Mathematics Question on Relations and functions
Find the range of each of the following functions.
(i) f(x) = 2 - 3x, x ∈ R, x> 0.
(ii) f(x) = x2+ 2, x, is a real number.
(iii) f(x) = x, x is a real number
(i) f(x) = 2 -3x, x ∈ R, x> 0
The values of f(x) for various values of real numbers x> 0 can be written in the tabular form as
x | 0.01 | 0.1 | 0.9 | 1 | 2 | 2.5 | 4 | 5 | ... |
---|---|---|---|---|---|---|---|---|---|
f(x) | 1.97 | 1.7 | -0.7 | -1 | -4 | -5.5 | -10 | -13 | ... |
Thus, it can be clearly observed that the range of fis the set of all real numbers less than 2.
i.e., range of f= (-, 2)
Alter:
Let x > 0
⇒3x > 0
⇒ 2-3x< 2
⇒ f(x) < 2
∴Range of f = (-, 2)
(ii) f(x) = x2+ 2, x, is a real number
The values of f(x) for various values of real numbers xcan be written in the tabular form as
x | 0 | ±0.3 | ±0.8 | ±1 | ±2 | ±3 | ..… | |
---|---|---|---|---|---|---|---|---|
f(x) | 2 | 2.09 | 2.64 | 3 | 6 | 11 | ..… |
Thus, it can be clearly observed that the range of fis the set of all real numbers greater than 2.
i.e., range of f= [2, ∞)
Alter:
Let x be any real number.
Accordingly,
x2 ≥0
⇒ x2+ 2 ≥0 + 2
⇒ x2+ 2 ≥2
⇒ f(x) ≥2
∴ Range of f = [2, )
(iii) f(x) = x, x is a real number
It is clear that the range of fis the set of all real numbers.
∴ Range of f = R