range of x+1/x

[apple2357]

I am looking at finding the range of the f(x)= x+1/x

One obvious way is to use calculus:

f'(x) = 1-1/x^2 , find the turning points and we get the range is f(x)>=2 or f(x)<=-2

Does the following alternative method make sense?

x+1/x to be rewritten as (sqrt(x)- i/sqrt(x) )^2 +2 , a sort of completing the square?
And whilst this gives one bit of the range ( defined only for x>0) how do I argue for the other bit?

 

[lex]

x+1/x to be rewritten as (sqrt(x)- i/sqrt(x) )^2 +2

[imath]\left(\sqrt{x}-\tfrac{i}{\sqrt{x}}\right)^2+2 =x-\tfrac{1}{x} - 2i + 2[/imath]

You might note that [imath]\left(\sqrt{x}-\tfrac{1}{\sqrt{x}}\right)^2+2 =x+\tfrac{1}{x} [/imath] and for x>0, the min of 2 is achieved when x=1 and there is no max.
You might also note that for [imath]f(x)=x+\tfrac{1}{x}, \hspace3ex f(-x) = -x+\tfrac{1}{-x} = -f(x)[/imath], so for x<0, when x=-1 the max of -2 is achieved and there is no min.

 

[apple2357]

[imath]\left(\sqrt{x}-\tfrac{i}{\sqrt{x}}\right)^2+2 =x-\tfrac{1}{x} - 2i + 2[/imath]

You might note that [imath]\left(\sqrt{x}-\tfrac{1}{\sqrt{x}}\right)^2+2 =x+\tfrac{1}{x} [/imath] and for x>0, the min of 2 is achieved when x=1 and there is no max.
You might also note that for [imath]f(x)=x+\tfrac{1}{x}, \hspace3ex f(-x) = -x+\tfrac{1}{-x} = -f(x)[/imath], so for x<0, when x=-1 the max of -2 is achieved and there is no min.

So we can say f(x) is an odd function which allows us to use rotational symmetry to see that it must also be true that x=-1 gives a max of -2 if its the case when x=1 we get a minimum of 2?

 

[lex]

Correct. Nice method!