# Math Help - maximum and minimum of functions

1. ## maximum and minimum of functions

calculate for each function:

a)the range of growth and decreasing
b)local extreme points
c)points of inflection
d)intervals of concavity
e)outline of the graphic

$1) y=xe^x$

$2) y=2\sqrt{x}-x$

$3)y=\frac{e^x}{x}$

2. Originally Posted by Apprentice123
calculate for each function:

a)the range of growth and decreasing
b)local extreme points
c)points of inflection
d)intervals of concavity
e)outline of the graphic

$1) y=xe^x$

$2) y=2\sqrt{x}-x$

$3)y=\frac{e^x}{x}$

I will give you a general outline as to how to do these problems.

Let $f(x)$ be twice differentiable function who we wish to find the max, mins, etc. of.

Max and mins

The method of finding max and mins is as follows.

First find $f'(x)$. After you have done so, find all x such that

$f'(x)=0$

Now consider these possible candidates for max or mins

Now you must do two things.

Firstly find $f''(x)$

Now test each x that made the first derivative zero in the second derivative. Now lets call that number a, if

$f''(a)<0\quad\text{This implies a relative maximum}$

If

$f''(a)>0\quad\text{This implies a relative minimum}$

If

$f''(a)=0\quad\text{This tells us nothing}$

If you do find values that make the second derivative zero, what you must do is subdivide the reals into subsets, the endpoints of each are the numbers that made the first derivative zero. So you should get something like this

$(-\infty,a)\cdots(b,c)(c,d)\cdots(e,\infty)$

Where a,b,c,d,e... were points that made the first derivative zero (there musn't be that many)

Now once we set this up we only want to check the value that made the second derivative zero, since this is a pain to do. So lets say that c was that pesky number such that

$f''(c)=0$

So we look at the intervals containing c, namely

$(b,c)(c,d)$

and what we do is test an x an element of each interval, i.e.

$f'\left(x\in(b,c)\right)\quad\text{and}\quad{f'\le ft(x\in(c,d)\right)}$

Now if there is a sign change between the two values there is a relative extrema there.

If it goes from plus to minus c is a relative max, and minus to plus c is a relative min.

And thats that

Inflection points

I won't got as in depth here, but what you must do is this

Find all x such that $f''(x)=0$ then take these values and do similarly to what we did with the intervals for the max or mins.

But this time, if there is any sign change (i.e. plus to minus OR minus to plus) that value is an inflection point.

Intervals of concavity

This is simply the subintervals of the reals we set up for the second derivative, the positive ones are intervals of positive concavity and obviously the negative ones are of negative concavity

When I say negative or positve I mean

$f''\left(\text{any }x\text{ an element of the interval}\right)=\pm$

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From there you ought to be able to draw a pretty good graph.

Hope this has helped

Mathstud.

3. thanks