Question

Suppose that f(x) is O (g(x)). Does it follow that 2^f(x) is O (2^g(x))?

Short answer

If f(x) is O (g(x)), then 2^f(x) is not O (2^g(x))

Step-by-step solution

Step 1

For example, f(x)=2x and g(x)=x

Now by the definition of Big-O notation,

$\underset{x\to \infty }{\lim }\frac{f\left(x\right)}{g\left(x\right)}$=$\underset{x\to \infty }{\lim }\frac{2x}{x}$

=constant< ∞

Thus f(x) is O (g(x)) is proved

Step 2

To check if 2^f(x) is O (2^g(x)),

Let h=2^2x and i(x)=2^x

Now by the definition of Big-O notation,

$\underset{x\to i}{\lim }\frac{h\left(x\right)}{i\left(x\right)}$ =C<∞to h=2^2x and i(x)=2^x

Step 3

$\begin{array}{r}=\underset{x\to \mathrm{\infty }}{lim} 2^{2x-x}\\ =\underset{x\to \mathrm{\infty }}{lim} 2^x\\ =\mathrm{\infty }\\ \end{array}$

2^f(x) is not O (2^g(x))

Final answer

If f(x) is O (g(x)), then 2^f(x) is not O (2^g(x))