Finding relative extrema

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lamaclass

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Hi I had a few uncertainties about these two problems:

1. Find all relative extrema. Use the Second Derivative Test where applicable.

f(x) = x + x/4

f'(x) = 1 - 4/X[sup:asyjzkim]2[/sup:asyjzkim]

f''(x) = 8/X[sup:asyjzkim]3[/sup:asyjzkim]

My book says that the critical points are 2 and -2 but I was wondering why zero would not be a critical point as well. Is it becuase when using the second derivative test, the test would fail so zero would not be counted as a critical point? And also, if that's the case, to determine where extrema are, would you apply the second derivative test to -2 and 2?

2. y = 1/2X[sup:asyjzkim]2[/sup:asyjzkim]-ln X

y' = X - 1/X

y'' = 1 + X/X[sup:asyjzkim]2[/sup:asyjzkim] > 0

So I know for this one there is a discontinuity at X = 1 but did not know if it had to applied to the first or second derivative tests to determine where the extrema are. Any help is greatly appreciated! :D
 
lamaclass said:
Hi I had a few uncertainties about these two problems:

1. Find all relative extrema. Use the Second Derivative Test where applicable.

f(x) = x + 4/x

f'(x) = 1 - 4/X[sup:nugh6wur]2[/sup:nugh6wur]

f''(x) = 8/X[sup:nugh6wur]3[/sup:nugh6wur]

My book says that the critical points are 2 and -2 but I was wondering why zero would not be a critical point as well.

for f'(x) = 0 ? x = ± 2

f(x) & f'(x) becomes undefined at x = 0, however for determining "relative" max/min, these points are excluded. Your books explanation is incomplete.
[edit]
A function has critical points at all points x[sub:nugh6wur]0[/sub:nugh6wur]where f'(x[sub:nugh6wur]o[/sub:nugh6wur]) = 0 or is f(x[sub:nugh6wur]o[/sub:nugh6wur]) not differentiable.
A real function is said to be differentiable at a point if its derivative exists at that point.
The given function is not differentiable at x = 0 , because lim h[sup:nugh6wur]+[/sup:nugh6wur]does not equal to lim h[sup:nugh6wur]-[/sup:nugh6wur].

Hence x = 0 is a critical point. However as stated above, for relative max/min it is not included in consideration.

Is it becuase when using the second derivative test, the test would fail so zero would not be counted as a critical point? And also, if that's the case, to determine where extrema are, would you apply the second derivative test to -2 and 2?

2. y = 1/2X[sup:nugh6wur]2[/sup:nugh6wur]-ln X

y' = X - 1/X

y'' = 1 + X/X[sup:nugh6wur]2[/sup:nugh6wur] > 0

So I know for this one there is a discontinuity at X = 1 but did not know if it had to applied to the first or second derivative tests to determine where the extrema are. Any help is greatly appreciated! :D
Click to expand...
 
\(\displaystyle First \ one: \ f(x) \ = \ x+\frac{4}{x}.\)

\(\displaystyle Definition \ of \ critical \ point, \ see \ below,\)

\(\displaystyle If \ f \ is \ defined \ at \ 0, \ then \ 0 \ is \ called \ a \ critical \ point \ if \ f' \ is \ undefined \ at \ 0.\)

\(\displaystyle Now, \ f(0) \ is \ undefined, \ ergo \ 0 \ is \ not \ a \ critical \ point.\)
 
A function has critical points at all points x[sub:1uwvhdv4]0[/sub:1uwvhdv4]where f'(x[sub:1uwvhdv4]o[/sub:1uwvhdv4]) = 0 or is f(x[sub:1uwvhdv4]o[/sub:1uwvhdv4]) not differentiable.

A real function is said to be differentiable at a point if its derivative exists at that point.

The given function is not differentiable at x = 0 , because lim h[sup:1uwvhdv4]+[/sup:1uwvhdv4]does not equal to lim h[sup:1uwvhdv4]-[/sup:1uwvhdv4] ? the derivative does not exist.

Hence x = 0 is a critical point.

However as stated before, for relative max/min, it is not included in consideration.[/color]
 
\(\displaystyle x \ = \ 0 \ is \ a \ vertical \ aysmptote, \ not \ a \ critical \ number.\)

\(\displaystyle For \ 0 \ to \ be \ a \ critical \ point \ it \ must \ be \ defined \ at \ f, \ which \ it \ is \ not.\)

\(\displaystyle f(0) \ = \ 0+\frac{4}{0}, \ is \ undefined. \ See \ definition \ above.\)

\(\displaystyle Definition \ of \ Critical \ Point:\)

\(\displaystyle If \ f \ is \ defined \ at \ c \ (0 \ being \ c \ in \ this \ case), \ then \ c \ is \ called \ a \ critical \ point \ of \ f \ if\)

\(\displaystyle f'(c) \ = \ 0 \ or \ if \ f' \ is \ undefined \ at \ c.\)

\(\displaystyle The \ sticking \ point: \ If \ f \ is \ defined \ at \ 0 \ - \ which \ it \ is \ not, \ hence \ 0 \ is \ not \ a \ critical \ point.\)

\(\displaystyle To \ wit, \ how \ can \ you \ have \ a \ critical \ point \ of \ something \ which \ is \ undefined \ at \ that \ point?\)

\(\displaystyle Impossible.\)
 
\(\displaystyle Not \ recieving \ a \ retraction \ from \ Subhotosh \ Khan, \ I \ wish \ to \ set \ the \ record \ straight \ on \\)

\(\displaystyle lamaclass \ inquiry, \ to \ wit: \ "My \ book \ says \ that \ the \ critical \ points \ are \ 2 \ and \ -2 \ but \\)

\(\displaystyle I \ was \ wondering \ why \ zero \ would \ not \ be \ a \ critical \ point \ as \ well."\)

\(\displaystyle Using \ a \ different \ approach: \ f(x) \ = \ x+\frac{4}{x} \ \implies \ domain \ = \ (-\infty,0)U(0,\infty)\)

\(\displaystyle Ergo, \ 0 \ is \ not \ in \ the \ domain \ of \ f \ (it \ doesn't \ exist \ in \ f), \ therefore \ it \ can't \ be \ a \ "critical\)

\(\displaystyle \ point" \ of \ f. \ I \ rest \ my \ case.\)

\(\displaystyle Addendum: \ You \ book \ is \ correct \ and \ is \ not \ incomplete.\)
 
Thanks a ton guys! Your explanations make a lot of sense and I get why now zero was not considered a critical point in that problem. :)
 
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