Question

Let f(x)=((x-1)^n)(x-2), where is an integer, n>=2.
For which values of n will x=1 gives:
1. a relative maximum?
2. a relative minimum?
3. neither

Answer 1

i think \(n=2\) will give a relative max

Answer 2

that's what i think (from the guess and check trials)
but.. since my teacher does not give us an interval range... i don't know how to calculate the minimum though

Answer 3

i don't think it asks for one, just for the \(n\)

Answer 4

if you make \(n=3\) the derivative will have a factor of \((x-1)^2\)which does not change sign at \(x=1\)

Answer 5

so that is a good candidate for "neither"

Answer 6

okay, but to find a relative minimum, shouldn't i calculate from a closed intervals?

Answer 7

relative is relative
for a closed interval you can find the absolute max and min, but on an open interval (like the whole real line) you can still compute a relative max and min

Answer 8

okay, how could i approach to the relative maximum then?
I know that plugging n=2 in the f'(x)=0 will give me 2 critical values, and from there i know the relative min... but how about relative max?

Answer 9

\(n=2\) gives a relative max
the relative max is 0

Answer 10

We know that the expansion of these functions for any n greater than or equal to 2 will give us a polynomial, which will always be differentiable at every point thus continuous at every point. Local extrema occur where the derivative of the function is 0 or the function is non-differentiable (But we only look for where it is 0 since polynomials are differentiable everywhere). These will be our critical points. To find these, we first differentiate the function. \[f'(x)=n(x-1)^{n-1}(x-2)+(x-1)^n|n \ge 2\]This first derivative tells us for x = 1 and any value of n greater than or equal to 2, there is always an extremum at x = 1. To find whether the extremum we found is a maximum or a minimum, we find the second derivative. Whereever the second derivative is positive, it's a minimum, and if it's negative, then it's a maximum.\[f''(x)=[n(n-1)(x-1)^{n-2}(x-2)+n(x-1)^{n-1}]+n(x-1)^{n-1}|n \ge 2\]Evaluating the second derivative for the first few values of n and plugging in x = 1, we notice that for every value of n after 2, the second derivative is 0 at x = 1. This means, that for any n > 2, there is neither a minimum or maximum at x = 1.

From this we conclude that for n = 2, at x = 1 there is a relative maximum. And for any n > 2, there exists neither a maximum or a minimum at x = 1.

@vivifang
@satellite73

Answer 11

@genius12 one minor detail, did you just optimize for "n" or for "x"?

Answer 12

I didn't optimize?

Answer 13

it says, optimize f(x=1)

Answer 14

for some "n"

Answer 15

Scroll up to the question. It's not asking me to optimise. Just asks me to find what type of extrema occurs at x = 1 for any n >= 2 in the given expression.

Answer 16

I do want to give you two (genius and satellite) as best responses... but apparently i can choose one >"<

Answer 17

You can fan me/write me a testimonial if you want @vivifang =D

Answer 18

which values of "n" would give,
1) a relative maxima
2) relative minima
3) neither..

Answer 19

wolfram
http://www.wolframalpha.com/input/?i=relative+extrema+f%28x%29%3D%28x-1%29^3+%28x-2%29

Answer 20

there is a minima for n=3

Answer 21

infasct, for odd "n", you have only a relative minima
for even "n" you have both relative maxima and mina

Answer 22

@electrokid The second derivative says it's neither concave up nor down for any n > 3. This means that's it's neither a maximum or a minimum. This becomes more evident as you increase the value of n. If you were to get really accurate, you could see that the values for x < 1 and x > 1 but very close to one are actually not the same as the value at x = 1 but the difference between them is next to nothing. From that we would be able to say that this is a minimum or a maximum. But in a general, not extremely picky conclusion through first/second derivatives you can conclude that it's neither for any n > 2 at x = 1.

Graphing this function for n tells you exactly what the story here is and how it's changing.
@electrokid

Answer 23

so after letting n=2, which function should i plug in? f(x)? f'(x)? or f''(x
)?
@ genius12

Answer 24

@genius12 so after determining n=12, which function should i plug in n=2?

Answer 25

What do you mean? @vivifang

Answer 26

i mean.. after determining n=2 (sorry for typo)

Answer 27

@genius12

Answer 28

Second one, f''(x).

Answer 29

But one very important thing that I just found @vivifang

Answer 30

Which is...? @genius12

Answer 31

When you start graphing like each these functions for different n's, you notice something. For values of x smaller than or greater than 1, but very close to 1, for all values of n, the function is always 0 for any x value very close to 1. This implies that if we are looking for relative extrema, then x =1 could be said to be both a minimum or a maximum even tho it has no concavity over an interval that goes from values less than but very close to one to values greater than 1 but also very close to 1.

For every n that is odd, we find something similar. If we look over an interval of values less than and greater than 1 but close 1, all of these values are 0 and so x = 1 could be said to be both a maximum and a minimum. But when you extend this interval to be rather farther away from 1, that is, instead of having the endpoints close to 1, make them slightly farther away and x = 1 could be characterised as having neither a maximum nor a minimum.

For even values of n, we find the same thing as for odd values of n. The only difference is, as you make the interval larger in which x = 1 is included and move the endpoints slightly farther away, we come to realise that x = 1 is a maximum.

From this we find that that x = 1, can be a maximum/minimum or neither for n values that are odd. And for n values that are even, x = 1 is only either a maximum/minimum.

This will make a lot more sense if you used a graphing calculator such as desmos.com and tried different values of n in the grapher and look at x = 1 to see what I mean.

@vivifang

Answer 32

so.. if i want to conclude the answer: "For even values of n, we find the same thing as for odd values of n. The only difference is, as you make the interval larger in which x = 1 is included and move the endpoints slightly farther away, we come to realise that x = 1 is a maximum.

From this we find that that x = 1, can be a maximum/minimum or neither for n values that are odd. And for n values that are even, x = 1 is only either a maximum/minimum." ?

Answer 33

Exactly.

Answer 34

But remember, n = 2 is an exception. Because that is a relative maximum. We are only talking above odd/even values of n > 2.

Answer 35

@vivifang

Answer 36

oh okay

Answer 37

thank you so much. <3 (this is the only question i can't solve for my Cal 1 assignment which is due tomorrow :P)