critical numbers question

[mathhelp1a]

1 / (1-x) ^ 2

is the critical number 1
and there is no inflection ?

 

[chrisr]

yes, there's no inflexion.
is that the function or derivative?
If it's the derivative of 1/(1-x),
there is an asymptote at x=1, graph goes to plus and minus infinity.
The slope is always positive for x not equal to 1.
f[sup:3abolrvg]'[/sup:3abolrvg](x) is never zero.

You should study the shape of the graphs,
it gets really clear then,
you will be able to see them in your mind's eye after.

 

[mathhelp1a]

no critical number and no relative extrema right

 

[chrisr]

the critical number may refer to a point at which the graph is "non-differentiable" also.
That would be at x=1, but there are no maxima or minima,
graph goes to infinity at x=1, split in two.
Check your class' usage of the word "critical".

 

[chrisr]

[attachment=0:1x9khfgz]Untitled.jpg[/attachment:1x9khfgz]

 

[BigGlenntheHeavy]

Here we go again.

$\displaystyle I \ am \ assuming \ that \ f(x) \ = \ \frac{1}{(1-x)^{2}}, \ hence \ f'(x) \ = \ \frac{2}{(1-x)^{3}}$

$\displaystyle Ergo, \ there \ are \ no \ critical \ numbers, \ including \ 1.$

DEFINITION OF CRITICAL NUMBER:

If f(1) is defined at 1, (which it isn't), then 1 is a critical number of f(x) if f'(1) is undefined.

Now f'(1) is undefined, but so is f(1), hence no critical number.

1 is not in the domain of f, hence how could it possibly be a critical number of f.

 

[BigGlenntheHeavy]

the critical number may refer to a point at which the graph is "non-differentiable" also.
That would be at x=1, but there are no maxima or minima,

The critical number may refer to a point at which the graph is "non-differentiable" , but the graph is continous.

The above graph is discontinous at x =1, hence no critical point.

$\displaystyle For \ example: \ Let \ f(x) \ = \ 2x-3x^{2/3}, \ then \ f'(x) \ = \ 2\bigg(\frac{x^{1/3}-1}{x^{1/3}}\bigg)$

$\displaystyle Critical \ numbers: \ x=1, \ and \ x=0.$

$\displaystyle Now, at \ x \ = \ 0, \ f \ is \ non-differentiable, \ but \ f(0) \ is \ still \ defined \ (continous), \ f(0) \ = \ 0$

$\displaystyle See \ graph.$

[attachment=0:386cjadv]pqr.jpg[/attachment:386cjadv]

 

[chrisr]

.

 

[BigGlenntheHeavy]

I wish I could squash this. If some one gives you a function and says y =f(x), find all critical numbers of f(x), then those numbers must be in the domain of f(x). If the number is not in the domain of f(x), it can't possibly be a critical number of f(x), that simple.