Let $f : R \to R$ be such that $f (x) = 2^{x}$ . Determine
(i) Range of $f$ (ii) ${x : f (x) = 1}$
(iii) whether $f (x + y) = f (x) f (y)$ holds.

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Solution

(I) $f (x) = 2^{x}$ as $x$ can take all real values, the range of $f (x)$ is $R^{+}$ or $(0, \infty)$
(ii) ${x : f (x) = 1}$
$\Rightarrow f (x) = 1$
$\Rightarrow 2^{x} = 1 = 2^{0} \Rightarrow x = 0$
(iii) $f (x + y) = 2^{x + y} = 2^{x} \times 2^{y} = f (x) f (y)$

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Let f:R→R be such that f(x)=2x. Determine(i) Range of f (ii) {x:f(x)=1}(iii) whether f(x+y)=f(x)f(y) holds.

(I) f(x)=2x as x can take all real values, the range of f(x) is R+ or (0,∞)(ii) {x:f(x)=1}⇒f(x)=1⇒2x=1=20⇒x=0(iii) f(x+y)=2x+y=2x×2y=f(x)f(y)

Let $f : R \to R$ be such that $f (x) = 2^{x}$ . Determine
(i) Range of $f$ (ii) ${x : f (x) = 1}$
(iii) whether $f (x + y) = f (x) f (y)$ holds.

(I) $f (x) = 2^{x}$ as $x$ can take all real values, the range of $f (x)$ is $R^{+}$ or $(0, \infty)$
(ii) ${x : f (x) = 1}$
$\Rightarrow f (x) = 1$
$\Rightarrow 2^{x} = 1 = 2^{0} \Rightarrow x = 0$
(iii) $f (x + y) = 2^{x + y} = 2^{x} \times 2^{y} = f (x) f (y)$