568 views

A superadditive function $f(\cdot)$ satisfies the following property $$f\left ( x_{1} +x_{2}\right )\geq f\left ( x_{1} \right ) + f\left ( x_{2} \right )$$

Which of the following functions is a superadditive function for $x > 1$?

1. $e^{x}$
2. $\sqrt{x}$
3. $1/x$
4. $e^{-x}$

Given that, a superadditive function $f(\cdot):$ $$f\left ( x_{1} +x_{2}\right )\geq f\left ( x_{1} \right ) + f\left ( x_{2} \right )$$

Since $x >1,$ a superadditive function can never be a decreasing function. So, optionc C and D can staright away be ruled out.

We can check options A and B by taking the value of $x_{1} = 2$ and $x_{2} = 3.$

1. $f(x) = \sqrt{x}$
$f(2+3) \geq f(2) + f(3)$
$\implies f(5) \geq f(2) + f(3)$
$\implies \sqrt{5} \geq \sqrt{2} + \sqrt{3}$
$\implies 2.236 \geq 1.414 + 1.732$
$\implies 2.236 \geq 3.146 \;{\color{Red} {\textbf{(False)}}}$

So, correct answer should be $(A).$

by
3.9k points

Solve it using exploring the options

Let $X_{1}$ = 2 and $X_{2}$ = 3

A. f(x) = $e^{x}$

$e^{5}$ > $e^{2}$ + $e^{3}$ ( using the calculator)

it satisfies the condition so option A is correct.

by
1.4k points

### 1 comment

Just one example working does not prove anything. A counter-example can prove a statement wrong but an example cannot prove a statement right.