# How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for  f(t) = 6t + 1/t?

May 5, 2017

$f \left(t\right) = 6 t + \frac{1}{t}$
$f ' \left(t\right) = 6 - \frac{1}{{t}^{2}}$
$f ' \left(t\right) > 0$ for $t > \frac{1}{\sqrt{6}} \mathmr{and} t < - \frac{1}{\sqrt{6}}$
$\therefore f \left(t\right)$ increases for $t > \frac{1}{\sqrt{6}} \mathmr{and} t < - \frac{1}{\sqrt{6}}$
And $f \left(t\right)$ decreases for $- \frac{1}{\sqrt{6}} <$$t$$<$$\frac{1}{\sqrt{6}}$
Now as the derivative changes from positive to negative at $t = - \frac{1}{\sqrt{6}}$ therefore it is a point of Maximum
Also as derivative changes from negative to positive at $t = \frac{1}{\sqrt{6}}$ therefore it is a point of minimum

May 5, 2017

The function is increasing when $x \in \left(- \infty , - \frac{1}{\sqrt{6}}\right) \cup \left(\frac{1}{\sqrt{6}} , + \infty\right)$
The function is decreasing when $x \in \left(- \frac{1}{\sqrt{6}} , 0\right) \cup \left(0 , \frac{1}{\sqrt{6}}\right)$
See below

#### Explanation:

Let's calculate the first derivative

$f \left(t\right) = 6 t + \frac{1}{t}$

The domain of $f \left(t\right)$ is ${D}_{f \left(t\right)} = \mathbb{R} - \left\{0\right\}$

$f ' \left(t\right) = 6 - \frac{1}{t} ^ 2$

The critical points are when $f ' \left(t\right) = 0$, that is

$6 - \frac{1}{t} ^ 2 = 0$

$6 = \frac{1}{t} ^ 2$

${t}^{2} = \frac{1}{6}$

$t = \pm \frac{1}{\sqrt{6}}$

Let's build a chart

$\textcolor{w h i t e}{a a a a}$$t$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- \frac{1}{\sqrt{6}}$$\textcolor{w h i t e}{a a a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a}$$\frac{1}{\sqrt{6}}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$f ' \left(t\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(t\right)$$\textcolor{w h i t e}{a a a a a a}$↗$\textcolor{w h i t e}{a a a a a a}$↘$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$↘$\textcolor{w h i t e}{a a a a}$↗

Let's calculate the second derivative

$f ' ' \left(t\right) = \frac{2}{t} ^ 3$

There is no point of inflection.

$f ' ' \left(- \frac{1}{\sqrt{6}}\right) = - 29.39$, as $f ' ' \left(- \frac{1}{\sqrt{6}}\right) < 0$, this a local maximum at $\left(- 0.408 , - 4.899\right)$

$f ' ' \left(\frac{1}{\sqrt{6}}\right) = 29.39$, as $f ' ' \left(\frac{1}{\sqrt{6}}\right) > 0$, this a local minimum at $\left(0.408 , 4.899\right)$

graph{6x+1/x [-41.1, 41.08, -20.54, 20.57]}