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Existence of a real-valued function yielding doubly-inconclusive second derivative test

Second partial derivative test is inconclusiveDoes picking $C=0$ as a constant of integration result in a nominated anti-derivative?Reconstructing a function from its critical points and inflection pointsIf the second derivative of a function is zero, why is the second derivative test inconclusive?Finding One-Sided Limits Algebraically by Breaking Functions into Piecewise PartsHigher Order Derivative Tests in Multiple DimensionsAPICS Mathematics Contest 1999: Prove $sin^2(x+alpha)+sin^2(x+beta)-2cos(alpha-beta)sin(x+alpha)sin(x+beta)$ is a constant function of $x$Inconclusive second derivative test rigorous proofSecond Partial Derivative Test for a Matrix Valued FunctionKernel of linear mapping from infinitely differentiable function to their derivatives

0

$begingroup$

Currently, I'm teaching a course in Calculus 1 and I would like to find a "simple" (whatever that means) real-valued function $f(x)$ that satisfies all the following properties:

1. 
   $x=a$ and $x=b$ are both in $operatornamedomain(f)subseteqmathbbR$
   


2. 
   $f'(a)=0$ and $f'(b)=0$
   


3. 
   $f''(a)=0$ and $f''(b)$ does not exist


4. 
   $f(x)$ isn't a piecewise function.

Assuming the above task is doable and not-too-daunting, I would also appreciate it if

5. the critical numbers of $f$ are findable using simple algebra,

as no calculators or CAS are allowed.

As the title indicates: Such an example would show that a single function having two critical numbers can yield inconclusive second derivative test in different ways. I think that would be pretty cool!

I've tried working backwards by considering possible functions $f''$ and taking their anti-derivatives. Everything I've tried has led to functions which violate (5) in a horrible way, so any insight would be hugely appreciated!

Note: The only reason I'm asking for something non-piecewise is because students are really bad at talking about one-sided limits, let alone one-sided derivatives.

calculus functions derivatives

share|cite|improve this question

asked Mar 16 at 22:01

cstovercstover

1227

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It might be worth waiting until you get to the fundamental theorem of calculus. Otherwise it seems like the only options you have for a function which is not differentiable at a point in it's domain are power functions.
  $endgroup$
  – Ethan Alwaise
  Mar 17 at 1:29
  

  
  
  
  

add a comment | 

0

$begingroup$

Currently, I'm teaching a course in Calculus 1 and I would like to find a "simple" (whatever that means) real-valued function $f(x)$ that satisfies all the following properties:

1. 
   $x=a$ and $x=b$ are both in $operatornamedomain(f)subseteqmathbbR$
   


2. 
   $f'(a)=0$ and $f'(b)=0$
   


3. 
   $f''(a)=0$ and $f''(b)$ does not exist


4. 
   $f(x)$ isn't a piecewise function.

Assuming the above task is doable and not-too-daunting, I would also appreciate it if

5. the critical numbers of $f$ are findable using simple algebra,

as no calculators or CAS are allowed.

As the title indicates: Such an example would show that a single function having two critical numbers can yield inconclusive second derivative test in different ways. I think that would be pretty cool!

I've tried working backwards by considering possible functions $f''$ and taking their anti-derivatives. Everything I've tried has led to functions which violate (5) in a horrible way, so any insight would be hugely appreciated!

Note: The only reason I'm asking for something non-piecewise is because students are really bad at talking about one-sided limits, let alone one-sided derivatives.

calculus functions derivatives

share|cite|improve this question

asked Mar 16 at 22:01

cstovercstover

1227

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It might be worth waiting until you get to the fundamental theorem of calculus. Otherwise it seems like the only options you have for a function which is not differentiable at a point in it's domain are power functions.
  $endgroup$
  – Ethan Alwaise
  Mar 17 at 1:29
  

  
  
  
  

add a comment | 

$begingroup$

Currently, I'm teaching a course in Calculus 1 and I would like to find a "simple" (whatever that means) real-valued function $f(x)$ that satisfies all the following properties:

1. 
   $x=a$ and $x=b$ are both in $operatornamedomain(f)subseteqmathbbR$
   


2. 
   $f'(a)=0$ and $f'(b)=0$
   


3. 
   $f''(a)=0$ and $f''(b)$ does not exist


4. 
   $f(x)$ isn't a piecewise function.

Assuming the above task is doable and not-too-daunting, I would also appreciate it if

5. the critical numbers of $f$ are findable using simple algebra,

as no calculators or CAS are allowed.

As the title indicates: Such an example would show that a single function having two critical numbers can yield inconclusive second derivative test in different ways. I think that would be pretty cool!

I've tried working backwards by considering possible functions $f''$ and taking their anti-derivatives. Everything I've tried has led to functions which violate (5) in a horrible way, so any insight would be hugely appreciated!

Note: The only reason I'm asking for something non-piecewise is because students are really bad at talking about one-sided limits, let alone one-sided derivatives.

calculus functions derivatives

share|cite|improve this question

asked Mar 16 at 22:01

cstovercstover

1227

$endgroup$

share|cite|improve this question

asked Mar 16 at 22:01

cstovercstover

1227

share|cite|improve this question

asked Mar 16 at 22:01

cstovercstover

1227

asked Mar 16 at 22:01

cstovercstover

1227

asked Mar 16 at 22:01

cstovercstover

1227