Question

Let f : R → R : f(x) = 2^x . Find 

(i) Range (f) 

(ii) {x : f(x) = 1}. 

(iii) Find out whether f(x + y) = f(x). f(y) for all x, y ϵ R.

Answer 1

Given that f: R → R such that f(x) = 2^x 

To find: (i) Range of x 

Here, f(x) = 2^x is a positive real number for every x ∈ R because 2^x is positive for every x ∈ R. 

Moreover, for every positive real number x, ∃ log2^x ∈ R such that

f(log₂x) = 2^log2x

= x [ ∵  a^logax = x]

Hence, the range of f is the set of all positive real numbers. 

To find: (ii) {x : f(x) = 1} 

We have, f(x) = 1 …(a) 

and f(x) = 2^x …(b) 

From eq. (a) and (b), we get 

2^x = 1 

⇒ 2^x = 2⁰ [∵ 2⁰ = 1] 

Comparing the powers of 2, we get 

⇒ x = 0 

∴{x : f(x) = 1} = {0} 

To find: (iii) f(x + y) = f(x). f(y) for all x, y ϵ R 

We have, 

f(x + y) = 2^{x + y} 

= 2^x .2^y 

[The exponent "product rule" tells us that, when multiplying two powers that have the same base, you can add the exponents or vice - versa] 

= f(x).f(y) [∵f(x) = 2^x ] 

∴ f(x + y) = f(x). f(y) holds for all x, y ϵ R