What is the relative minimum for y = x + 1/x

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- Jun 7th 2006, 11:13 PMnath_quamRelative Minimum
What is the relative minimum for y = x + 1/x

- Jun 8th 2006, 12:22 AMCaptainBlackQuote:

Originally 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 - Jun 8th 2006, 12:23 AMTD!
Solve dy/dx = 0, you'll find a relative min and max.