Can You Simplify the Second Derivative Test Explanation?

[Petrus]

This is a problem from My book which I have hard understanding what they are asking for, I am pretty confused on the question would like to have help!

Second derivate test works as follows:
If f (c) = 0 and f'' (c)> 0 Then c is a local min point for function f (a) Show that c need not be a local minimum of f if one removes the requirement "f'' (c)> 0
(but retains the other requirements), by giving an example of such a function.
(B) Show that c may be a local minimum even if you take away the requirement "f'' (c)> 0 (but retains the other requirements), by giving an example of such a function.
(Because the derivative in an extreme point according to Fermat's statement above is always zero if it is defined there is no point to look at what happens if you take away the requirement "f '(c) = 0.")
regards,
\(\displaystyle |\pi\rangle\)

 

[Ackbach]

This is a problem from My book which I have hard understanding what they are asking for, I am pretty confused on the question would like to have help!

Second derivate test works as follows:
If $f (c) = 0$ and $f'' (c)> 0$ Then $c$ is a local min point for function $f$

I think you meant "If $f'(c)=0$ and $f''(c)>0$, then $c$ is a local min for $f$. The actual value of the function is irrelevant.

(a) Show that $c$ need not be a local minimum of $f$ if one removes the requirement $f'' (c)> 0$.
(but retains the other requirements), by giving an example of such a function.

Can you think of a function where $f'(c)=0$ but $f''(c) \le 0$, and $c$ is not a local min?

(B) Show that $c$ may be a local minimum even if you take away the requirement $f'' (c)> 0$ (but retains the other requirements), by giving an example of such a function.
(Because the derivative in an extreme point according to Fermat's statement above is always zero if it is defined there is no point to look at what happens if you take away the requirement $f '(c) = 0.$)

regards,
\(\displaystyle |\pi\rangle\)

So here, here you need a function where $f'(c)=0$, and $f''(c) \le 0$, and $c$ is a local min. Can you think of one?

 

[Petrus]

I think you meant "If $f'(c)=0$ and $f''(c)>0$, then $c$ is a local min for $f$. The actual value of the function is irrelevant.
Can you think of a function where $f'(c)=0$ but $f''(c) \le 0$, and $c$ is not a local min?
So here, here you need a function where $f'(c)=0$, and $f''(c) \le 0$, and $c$ is a local min. Can you think of one?

(a) \(\displaystyle f(x)-x^2\)
(b) I got no clue of that one

 

[Fermat1]

This is a problem from My book which I have hard understanding what they are asking for, I am pretty confused on the question would like to have help!

Second derivate test works as follows:
If f (c) = 0 and f'' (c)> 0 Then c is a local min point for function f (a) Show that c need not be a local minimum of f if one removes the requirement "f'' (c)> 0
(but retains the other requirements), by giving an example of such a function.
(B) Show that c may be a local minimum even if you take away the requirement "f'' (c)> 0 (but retains the other requirements), by giving an example of such a function.
(Because the derivative in an extreme point according to Fermat's statement above is always zero if it is defined there is no point to look at what happens if you take away the requirement "f '(c) = 0.")
regards,
\(\displaystyle |\pi\rangle\)

Try and think what the graphs would look like in each case. For the first, think of a function with a maximum but no minmum. For example $f(x)=-x^2$ and $c=0$. For B), what you are showing is that this condition is sufficient, but not neccesary. As I understand it, we still want $f'(c)=0.$ What about the constant function $f=c$. Then $f'=f''=0$ but $c$ is still a minumum