# Stumped on precalc problem

1. Sep 21, 2011

### HarveyBullock

1. The problem statement, all variables and given/known data

1. given F(x) = .2x^4 - 2x^2 - 5x
find: XMIN, XMAX, YMIN, YMAX

2. Using F(X) find the following:
a. Domain
b. Range
c. List increasing intervals
d. List decreasing intervals
e. Coordinates of any maximums or minimums

I worked it out mostly on paper and can't get that far. I did figure out though that the function is neither odd nor even. which was for a different question

Last edited: Sep 21, 2011
2. Sep 21, 2011

### symbolipoint

Since you are not looking for roots but want minima and maxima and the problem is for Precalculus, easiest and fastest is to use a graphing calculator.

3. Sep 21, 2011

### HarveyBullock

I entered it on the calculator but did not really see the "hump" do figure out maxima and minima

4. Sep 22, 2011

### Allenman

Did you learn any rules for differentiation?

If you differentiate it, you're finding the equation of the slope of your original function.
Set the new equation equal to zero and solve for different values of x to get the critical points (don't forget the endpoints are also critical points).
Plug in numbers on either side of those points to find whether it's positive or negative (increasing or decreasing).... If the critical point is between a positive and a negative slope, it will be a local max or min (depending on which side has the positive and negative).

5. Sep 22, 2011

### symbolipoint

You will see an inflection, but no other kind of hump. The question is here in the Precalculus section of this category of the forum, so no calculus assumed and none expected. With a graphing program or tool, you will find two Real roots, and there is only one minimum; no other minimum or maximum.

For viewing, you could set to XMIN = ~-10, XMAX = ~+15, YMIN = -20, YMAX = ~+15.
The one obvious root is (0, 0), and the other root just a fraction of a unit greater than x=4 (but you find it yourself using your program or your graphing calculator).

Your function can be transformed into $F(x)=\frac{1}{5}x(x^3-10x-25)$. When you try examining the cubic factor with the Rational Roots method, you may find you cannot obtain another polynomial factor. I suspect that two factors for F(x) could contain either non-rational roots or complex roots.

6. Sep 22, 2011

### HallsofIvy

This makes no sense. First, obviously, there is NO y and so no "YMIN" or "YMAX"! I assume you mean y= F(x). More importantly, unless additional conditions are given, there is not limit in x and so there is no "XMIN" or "XMAX". (Perhaps they intend XMIN= -infinity and XMAX= +infinity.)

Are you sure about that "x" in "-5x"? Without it, we could let $u= x^2$, so the equation becomes $u= .2u^2- 2u- 5= (1/5)(u^2- 10u- 25)=(1/5)(u^2- 10u+ 25- 50)= (1/5)((u- 5)^2- =50)$. In that form, it is relatively easy to determine the values you want.

If it really is $F(x)= .2x^4- 2x^2- 5x= (1/5)x(x^3- 10x- 25)$. Now, you would have to use differentiation to find max and min.

7. Sep 22, 2011

### Swalker

Hi,

It is clear that f(x) will be plotted on an x, y graph so xmin, xmax correspond to the x axis minima and maxima humps, ymin and ymax correspond to the values.

To search for these using your graphing calculator stay with a large range like
Xmin & ymin = -50 and then xmax and ymax = 50.

Once you see humps on the graph zoom in to the ranges to get the actual values.

Your domain will be xmin, xmax.

Your range will be ymin to ymax.

Hope this helps