mathhelp1a
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1 / (1-x) ^ 2
is the critical number 1
and there is no inflection ?
is the critical number 1
and there is no inflection ?
critical numbers question
- Thread starter mathhelp1a
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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.
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
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no critical number and no relative extrema right
BigGlenntheHeavy
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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.
\(\displaystyle I \ am \ assuming \ that \ f(x) \ = \ \frac{1}{(1-x)^{2}}, \ hence \ f'(x) \ = \ \frac{2}{(1-x)^{3}}\)
\(\displaystyle Ergo, \ there \ are \ no \ critical \ numbers, \ including \ 1.\)
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
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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]
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]
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BigGlenntheHeavy
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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.