Second Derivative Test w/ no critical numbers

May 2010
2
0
Hello,

Here is a problem that I'm really struggling with: f(x)=2-2x-x^3. F prime of this function is -2-3x^2 and when setting to zero to find critical numbers, you end up taking the sq. root of -2/3. So there are no critical numbers.

I'm supposed to determine concavity and find inflection points. Per the second derivative test rules, F prime c must equal 0 and f double prime must exist near c. Since F prime C doesn't equal zero, as there are no critical numbers, is the graph also not concave up/down and are there also no inflection points? Am I getting this right?

I'm so confused, Thank you.
 

mr fantastic

MHF Hall of Fame
Dec 2007
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Hello,

Here is a problem that I'm really struggling with: f(x)=2-2x-x^3. F prime of this function is -2-3x^2 and when setting to zero to find critical numbers, you end up taking the sq. root of -2/3. So there are no critical numbers.

I'm supposed to determine concavity and find inflection points. Per the second derivative test rules, F prime c must equal 0 and f double prime must exist near c. Since F prime C doesn't equal zero, as there are no critical numbers, is the graph also not concave up/down and are there also no inflection points? Am I getting this right?

I'm so confused, Thank you.
Application of the definitions:

Concave up: f'' > 0.

Concave down: f'' < 0.

The function is a cubic and therefore has a point of inflection - the x-coordinate is therefore found by solving f'' = 0.
 
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May 2010
2
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So the fact that f prime has no real critical numbers is irrelevant to the second derivative test for determining concavity?

F double prime's critical point is (0,2) and is also the point of inflection. Is this correct?
 

mr fantastic

MHF Hall of Fame
Dec 2007
16,948
6,768
Zeitgeist
So the fact that f prime has no real critical numbers is irrelevant to the second derivative test for determining concavity?

F double prime's critical point is (0,2) and is also the point of inflection. Is this correct?
Yes. Yes. A simple sketch graph of f(x) = 2 - 2x - x^3 confirms all this: plot y =2-2x-x^3 - Wolfram|Alpha
 
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