# How do you find the local max and min for f(x) = 2 x + 3 x ^{ -1 } ?

Nov 20, 2016

Minimum=$x = \sqrt{\frac{3}{2}}$

#### Explanation:

Think about a line when it hits its maximum and then begins to curve back down. At its very maximum, its slope will be zero, and the same for the minimum. Therefore, what you must do is find the derivative and equal it to zero and solve, because the derivative is equal to the slope at a given time.

$f ' \left(x\right) = 2 - 3 {x}^{-} 2 = 0$
$3 {x}^{-} 2 = 2$
$3 = 2 {x}^{2}$
$x = \sqrt{\frac{3}{2}}$

By making the derivative equal to zero, you are finding one or more "critical points" that can either be a maximum of the function, minimum of the function, or neither. Because you don't know what these are, you can use the first derivative test.

In the first derivative test, you plug in a two values of x into the derivative equation, one greater and one lesser than the solution found. Let's use 1 and 2.
1
$2 - 3 {x}^{-} 2$
$2 - 3 {\left(1\right)}^{-} 2$
$2 - 3 = - 1$
This solution means that there is a NEGATIVE slope to the left of the point we found.
2
$2 - 3 {\left(2\right)}^{-} 2$
$2 - \frac{3}{4} = \frac{5}{4}$
This solution means that there is a POSITIVE slope to the right of the point we found.
This means that we have found a minimum. Visualize a dip in a graph- it goes down (negative slope) and then goes up again (positive slope).

We did not find any other values when we made the derivative equal to zero, so there are no maximums or other minimums in this graph!
If you want to find the coordinate point, just plug $x = \sqrt{\frac{3}{2}}$ into f(x).