absolute value question

[lizzpalmer]

F(x) = x^4 – 18x^2 + 1; [-4,4]

so from what I gather, we have to solve for 0

x^4 – 18x^2 + 1 = 0

x^4 – 18x^2 = -1

my issue is what do i do next? not sure how to solve further properly. If anyone could help?

 

[JeffM]

F(x) = x^4 – 18x^2 + 1; [-4,4]

so from what I gather, we have to solve for 0

x^4 – 18x^2 + 1 = 0

x^4 – 18x^2 = -1

my issue is what do i do next? not sure how to solve further properly. If anyone could help?

if that is the entire question, it is cryptic in the extreme so I do not know whether I am being helpful.

If the problem is about finding real roots (zeros) of F(x) in the interval [-4, 4], consider this hint:
Let u = x[sup:22vihd2e]2[/sup:22vihd2e].
Let F(x) = G(u).
What are the real roots of G(u)?

 

[mmm4444bot]

F(x) = x^4 – 18x^2 + 1; [-4,4]

so from what I gather, we have to solve for 0 Huh?

Does your on-line course provide instructions with each exercise? :? You should not have to guess.

In particular, I would like to know why you described this exercise as an "absolute value question" and why you posted the interval [-4, 4].

If you have any information that you did not post, will you please share it? Maybe we can figure out what they want you to do.

 

[lizzpalmer]

The question says:

Find the absolute extreme of each function given the interval.

We are supposed to use the extreme value theorem

The [-4,4} are the intervals.

The instructor has told me to solve for 0 but my problem is I can't seem to do it correctly. He says I need to find f'(x) first and I think I keep messing that up.

so if I have f(x) = x^4 - 18x^2 + 1

I thought I would do:

f ' (x) = 4x^3 - 36

then to solve the problem I would substitute for zero:

4x^3 - 36 = 0

4x^3 = 36

x^3 = 9

But I must have something wrong because I am not sure what to do next. I need to come up with the absolute minimum and absolute maximum.

 

[JeffM]

The question says:

Find the absolute extreme of each function given the interval. The extremes (extrema) are just the minimum and maximum values of f(x) within the given interval.

We are supposed to use the extreme value theorem The extreme value theorem says that if f(x) is real and continuous in the bounded interval, then both a minimum f(x) and a maximum f(x) for that interval exist. So is f(x) = x[sup:3uj8xgbk]4[/sup:3uj8xgbk] - 18x[sup:3uj8xgbk]2[/sup:3uj8xgbk] + 1 real and continuous over the interval [-4, 4]? If it is, the problem is soluble, at least in principle.

The [-4,4} are the intervals.

The instructor has told me to solve for 0 but my problem is I can't seem to do it correctly. He says I need to find f'(x) first and I think I keep messing that up.

so if I have f(x) = x^4 - 18x^2 + 1

I thought I would do:

f ' (x) = 4x^3 - 36 This is an error. What is the derivative of - 18x[sup:3uj8xgbk]2[/sup:3uj8xgbk]?

I need to come up with the absolute minimum and absolute maximum.

If f'(x) = 0 and f''(x) < 0 at x = a, f(a) is a LOCAL maximum, but it may not be a GLOBAL maximum. Similarly, if f'(x) = 0 and f''(x) < 0 at x = a, f(a) is a LOCAL minimum, but it may not be a GLOBAL minimum. Indeed, there may be multiple local extrema within an interval. So, when searching for the extrema in an interval, you need to look for all the local extrema within the interval, determine what f(x) equals at each local extremum and at the endpoints of the interval. Do you see why you need to look at the endpoints as well as every local extremum?

 

[mmm4444bot]

Find the absolute [extrema] of each function [in the given] interval.

Ah, this is why I was confused. The phrases "absolute value" and "absolute extreme" do not have the same meaning.

Hence, this exercise is not an absolute-value question.


The instructor has told me to solve for 0

Yes, but not the equation that you posted.

The instructor should have said something like, "solve the derivative for 0". That's another reason for my confusion.

As Jeff explained, the absolute extrema are the largest and the smallest value of f(x), when the domain of f(x) is restricted to [-4, 4].

A picture of function f's behavior over this interval would be a nice adjunct to using the derivative. Your TI-83 can draw this for you.

[attachment=1:32eaznev]quartic.JPG[/attachment:32eaznev]

We see that f(-4) and f(4) are not extreme values. Jeff explained that it's important to remember to check the function value at the endpoints of the interval. This is crucial, if you do not have a picture of the function's behavior. Look what happens when the interval is increased to [-4.5,4.5]. The largest value of f(x) now occurs at the endpoints of the interval, as opposed to some location where the derivative is zero.

[attachment=0:32eaznev]example.JPG[/attachment:32eaznev]

The concepts of "even functions" and "symmetry" are also nice to know, but not essential.

Cheers ~ Mark