# Maximum and minimum functions

• Jun 23rd 2008, 03:23 PM
Apprentice123
Maximum and minimum functions
To determine the role given to:
a) the range of growth and decrease
b) points of extreme local
c) points of inflection
d) intervals as the concavity
e) outline of the plot

function:

$\displaystyle \frac{x^2+1}{x}$

Sorry I do not have the answer
• Jun 23rd 2008, 05:56 PM
topsquark
Quote:

Originally Posted by Apprentice123
To determine the role given to:
a) the range of growth and decrease
b) points of extreme local
c) points of inflection
d) intervals as the concavity
e) outline of the plot

function:

$\displaystyle \frac{x^2+1}{x}$

Sorry I do not have the answer

Mostly this is basic definition work.
a) You can tell if a function is increasing or decreasing by looking at the sign of the first derivative. For example, your first derivative is
$\displaystyle \frac{(2x)x - (x^2 + 1) \cdot 1}{x^2} = \frac{x^2 - 1}{x^2}$
Where is this equal to 0?
$\displaystyle \frac{x^2 - 1}{x^2} = 0$

$\displaystyle x^2 - 1 = 0$

$\displaystyle x = \pm 1$

So divide the real line into three intevals:
$\displaystyle (-\infty, -1),~(-1, 1)~(1, \infty)$
and see where the derivative is positive and negative.

b) You already located the critical points when you found the first derivative and set it equal to 0.

c) You have an inflection point where the second derivative is equal to 0.

d) Do this the same way as part a), only use the second derivative and set it equal to 0. Where is the second derivative positive or negative?

e) The information above tells you a great deal about the curve. Two additional pieces of information will be helpful. Let
$\displaystyle y = \frac{x^2 + 1}{x}$

The y-intercept is the point on the graph such that x = 0.
The x-intercepts are the points on the graph where y = 0.

-Dan
• Jun 24th 2008, 05:47 AM
Apprentice123
thank you.