# Relative Minimum

• June 8th 2006, 12:13 AM
nath_quam
Relative Minimum
What is the relative minimum for y = x + 1/x
• June 8th 2006, 01:22 AM
CaptainBlack
Quote:

Originally Posted by nath_quam
What is the relative minimum for y = x + 1/x

A local minimum of a function can be found by differentiating the function
and solving for the zeros of the derivative. These zeros comprise local
maxima, minima and points of inflection, so you will need to confirm
which of these any such zeros is.

$
y=x+1/x
$
,

so:

$
\frac{dy}{dx}=1+(-1)/x^2
$

So we are looking for solutions of:

$
1-\frac{1}{x^2}=0
$

which with a bit of algebraic jiggery-pokery we find are:

$x=\pm 1$,

and a bit more work shows that $x=1$ is a local minimum,
and $x=-1$ is a local maximum.

RonL
• June 8th 2006, 01:23 AM
TD!
Solve dy/dx = 0, you'll find a relative min and max.