# How do you find all local maximum and minimum points using the second derivative test given y=x+1/x?

Oct 18, 2016

Please see the explanation section below

#### Explanation:

$y = f \left(x\right) = x + \frac{1}{x}$.

Find the critical numbers for $f$

The domain of $f$ is $\left(- \infty , 0\right) \cup \left(0 , \infty\right)$

$y ' = 1 - \frac{1}{x} ^ 2 = \frac{{x}^{2} - 1}{x}$

$y '$ fails to exist at $0$ which is not in the domain, so is not a critical number.

$y ' = 0$ at $- 1$ and at $1$. Both are in the domain, so both are critical numbers.

Find the second derivative

$y ' ' = f ' ' \left(x\right) = \frac{2}{x} ^ 3$

Apply the test

At $x = - 1$ we have $y ' ' = f ' ' \left(- 1\right) = \frac{2}{- 1} ^ 3 < 0$.

Since the second derivative is negative at the critical number $- 1$, we conclude that $f \left(- 1\right)$ is a relative maximum.

There is a relative maximum of $- 2$ at $x = - 1$.

I assume that "ralative maximum point" means "point on the graph where $y$ is a relative maximum". If so, the answer should be

Relative maximum point: $\left(- 1 , - 2\right)$

At $x = 1$ we have $y ' ' = f ' ' \left(1\right) = \frac{2}{1} ^ 3 > 0$.

Therefore $f \left(1\right) = 2$ is a relative minimum.

Again, I assume the requested form for the answer is

Relative minimum point: $\left(1.2\right)$

Here is the graph of $y = x + \frac{1}{x}$
Note the discontinuity at $0$ that results in two separate branches of the graph. One branch has a maximum of $- 2$ at $- 1$ and the other has a minimum of $2$ at $1$.
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