What is the relative minimum for y = x + 1/x
A local minimum of a function can be found by differentiating the functionOriginally Posted by nath_quam
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.
$\displaystyle
y=x+1/x
$,
so:
$\displaystyle
\frac{dy}{dx}=1+(-1)/x^2
$
So we are looking for solutions of:
$\displaystyle
1-\frac{1}{x^2}=0
$
which with a bit of algebraic jiggery-pokery we find are:
$\displaystyle x=\pm 1$,
and a bit more work shows that $\displaystyle x=1$ is a local minimum,
and $\displaystyle x=-1$ is a local maximum.
RonL