Find the absolute max/min value for f(x) = x + x^(-1), x > 0

[K_Swiss]

Find the absolute maximum and minimum values for:

$\displaystyle f(x) = x + x^{-1}, x > 0$

My work:

$\displaystyle f'(x) = 1 - \frac {1}{x}$

$\displaystyle 0 = \frac {x-1}{x}$

$\displaystyle x - 1 = 0$
$\displaystyle x = 1$

$\displaystyle f(1) = 1 + 1^{-1}$
$\displaystyle f(1) = 2$

Critical Point - $\displaystyle (1, 2)$

Since $\displaystyle x > 0$, can pick any x value greater than 0...

$\displaystyle x = 3$
$\displaystyle f(3) = 3 + 3^{-1}$
$\displaystyle f(3) = \frac {10}{3}$

Point$\displaystyle (3, \frac {10}{3})$

Therefore, the absolute maximum is at $\displaystyle x > 0$ and the absolute minimum is at $\displaystyle (1,2)$


Am I right, if not, please correct my work.

 

[galactus]

There is a slight mistake.

Your first derivative is $\displaystyle f'(x)=1-\frac{1}{x^{2}}$

This gives two solutions. A max at x=-1 and a min at x=1

Here's the graph:

 

[K_Swiss]

Would the critical points be (-1, -2) and (1, 2)?

I don't understand why x = -1 is the max? (1, 2) looks like the local max because of the greater y value? I know the graph proves otherwise, but I need to know how to prove it algebraically

 

[galactus]

Try the second derivative test.

If f''(x)>0 at some point, then f has a relative minimum.

If f''(x)<0 at some point, then f has a relative maximum.

$\displaystyle f''(x)=\frac{2}{x^{3}}$

$\displaystyle f''(1)=2$, therefore it has a relative minimum at x=1

$\displaystyle f''(-1)= -2$, therefore, it has a relative maximum at x=-1

 

[K_Swiss]

Thanks for the help