Function , Functions!

By AM-GM inequality we have, $x+\frac{1}{x} \geq 2\sqrt{x \times \frac{1}{x}}$ $\Rightarrow x+\frac{1}{x} \geq 2$

$\therefore$ the minimum value of $f(x)$ is $2$ , and the minimality is achieved when $x=\frac{1}{x}=1$ .

d/dx(f(x))=0

0=d/dx(x+1/x)

0=1-1/x^2

x^2=1

x=1

mininum value of f(x) = 1+1=(2)

It is important to state for real value for x and hence for f (x). There was same question that I didn't answer because the question didn't mention about real only.

If x could be non real numbers, then f (x) could choose to be any real values; plug in any big value for f (x) and solve for complex x. Further, only real numbers are having greater or smaller and hence maximum or minimum.

f' (x) = 1 - 1/ x^2

f'' (x) = 2/ x^3

f' (x) = 0 for x = - 1 or 1,

f'' (-1) < 0 for a maximum and f'' (1) > 0 for a minimum.

For only positive real number, 1+ 1/ 1 as minimum is obvious when we compare to 2 + 1/ 2 for example.

x + 1/ x = k => x^2 - k x + 1 = 0

(x - k/ 2)^2 = k^2/ 4 - 1 indicates that k mustn't be < 2 for real x, equal to say that k >= 2 or f (x) = x + 1/ x >= 2. Minimum value of f (x) is 2.