First Derivative Test for Finding Local Extremas of x^4 - 50x^2 + 4

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Homework Statement

Consider the equation below. (Round the answers to two decimal places. If you need to use -infinity or infinity, enter -INFINITY or INFINITY.)
f(x) = x^4 - 50x^2 + 4

Find the local minimum and maximum values of f.

Homework Equations

First derivative test and that is about it

The Attempt at a Solution

Take the derivative using the power rule so
f ' (x)= 4x^3-100x

Then to find critical points, set f' (x)=0 and solve for x

0=4x^3-100x
0=4x(x^2-25), x=-5,0,5

then plug those x values into the original function and you get

f(0)=4 ------max
f(-5)= -621-----min
f(5)=-621------min

This is what i get for an answer, however the where I can type these answers online there are only two boxes. One is for min and one is for max. What did i do wrong?

 

[n!kofeyn]

How are you determining those to be local maximums and minimums just based upon their value? I think you are confusing the first derivative test for local extrema and the first derivative test for absolute extrema on closed intervals. You are dealing with the entire real line and asked to find the local extrema.

Once you have found your critical points, that is x=-5,0,5, you break up the real line into intervals where you exclude these points. So you will get (-∞,-5), (-5,0), (0,5), and (5,∞). Then choose a point in each interval and evaluate f'(x) at that point. If f' is positive at that point, then the function f is increasing on that interval, if it is negative, then f is decreasing on that interval.

If to the left of a critical point, the function is increasing, and to the right, the function is decreasing, then that point gives you a local maximum. You carry on in this way, until you have categorized each critical point as giving either a local maximum, minimum, or neither. Evaluate f at those points to actually get the maximum/minimum values.

 

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But isn't that what i did?

 

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There are three critical points. By seeing what intervals of the four go up and down, I believe you get the same answer i do. No?

 

[n!kofeyn]

No it isn't. Did you read what I wrote? You just evaluated f at the critical points, and that was it. You got lucky that in this problem all three critical points gave extrema, but that isn't always going to be the case. For example, look at the following http://www.wolframalpha.com/input/?i=plot+x^4-2x^3-2". Notice that f'(x)=0 at x=0, but it is neither a maximum nor a minimum.

By your method above, you would have gotten that x=0 gave a maximum value simply because it is larger than the value of the function evaluated at the other critical point.

 

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I still get the same answer using your method. Intervals that are increasing include (-5,0),(5,infinity) and decreasing intervals are (-infinity,-5),(0,5).

"If to the left of a critical point, the function is increasing, and to the right, the function is decreasing, then that point gives you a local maximum. You carry on in this way, until you have categorized each critical point as giving either a local maximum, minimum, or neither. Evaluate f at those points to actually get the maximum/minimum values."

going through it decreases on -infinity to -5 and then increases from - 5 to 0. local min. -5 to 0 it increases and 0 to 5 it decreases. local max. 0 to 5 it decreases and 5 to infinity it increases. local min.

 

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i still don't understand what I am doing wrong so

 

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nevermind. i figured it out.

 

[n!kofeyn]

Yes, you got the same answer because you got lucky. You're method was just wrong. Also, the answer only has two answer boxes because they are asking for the maximum and minimum values.

 

[hover]

I understand now. I learned the correct way from you on how to do it. The value thing tripped me up. thanks for the help.

 

[n!kofeyn]

No problem!