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.

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 , ∃ log_{2}x ∈ R such that

= 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} = 2^{0} [∵ 2^{0} = 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

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