Differentiable function with no second derivative at $0$?












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What is an example of a function that is differentiable, concave up everywhere, and $f''(0)$ does not exist?






calculus analysis examples-counterexamples







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  • 3




    $begingroup$
    $f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
    $endgroup$
    – Michael Hoppe
    Dec 6 '18 at 16:57










  • $begingroup$
    @Eric Maybe only integral of absolute value?
    $endgroup$
    – Юрій Ярош
    Dec 6 '18 at 17:01










  • $begingroup$
    @ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
    $endgroup$
    – Eric
    Dec 6 '18 at 17:04


















0












$begingroup$


What is an example of a function that is differentiable, concave up everywhere, and $f''(0)$ does not exist?






calculus analysis examples-counterexamples







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
    $endgroup$
    – Michael Hoppe
    Dec 6 '18 at 16:57










  • $begingroup$
    @Eric Maybe only integral of absolute value?
    $endgroup$
    – Юрій Ярош
    Dec 6 '18 at 17:01










  • $begingroup$
    @ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
    $endgroup$
    – Eric
    Dec 6 '18 at 17:04
















0












0








0





$begingroup$


What is an example of a function that is differentiable, concave up everywhere, and $f''(0)$ does not exist?






calculus analysis examples-counterexamples







share|cite|improve this question











$endgroup$




What is an example of a function that is differentiable, concave up everywhere, and $f''(0)$ does not exist?






calculus analysis examples-counterexamples




calculus analysis examples-counterexamples






share|cite|improve this question















share|cite|improve this question













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edited Dec 6 '18 at 16:59









Andrews

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3901317










asked Dec 6 '18 at 16:49









Y.Z.Y.Z.

153




153








  • 3




    $begingroup$
    $f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
    $endgroup$
    – Michael Hoppe
    Dec 6 '18 at 16:57










  • $begingroup$
    @Eric Maybe only integral of absolute value?
    $endgroup$
    – Юрій Ярош
    Dec 6 '18 at 17:01










  • $begingroup$
    @ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
    $endgroup$
    – Eric
    Dec 6 '18 at 17:04
















  • 3




    $begingroup$
    $f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
    $endgroup$
    – Michael Hoppe
    Dec 6 '18 at 16:57










  • $begingroup$
    @Eric Maybe only integral of absolute value?
    $endgroup$
    – Юрій Ярош
    Dec 6 '18 at 17:01










  • $begingroup$
    @ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
    $endgroup$
    – Eric
    Dec 6 '18 at 17:04










3




3




$begingroup$
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
$endgroup$
– Michael Hoppe
Dec 6 '18 at 16:57




$begingroup$
$f(x)=x^2$ for $x<0$ and $f(x)=x^3$ else.
$endgroup$
– Michael Hoppe
Dec 6 '18 at 16:57












$begingroup$
@Eric Maybe only integral of absolute value?
$endgroup$
– Юрій Ярош
Dec 6 '18 at 17:01




$begingroup$
@Eric Maybe only integral of absolute value?
$endgroup$
– Юрій Ярош
Dec 6 '18 at 17:01












$begingroup$
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
$endgroup$
– Eric
Dec 6 '18 at 17:04






$begingroup$
@ЮрійЯрош No, I just messed up. The first integral isn't concave up. You can take the first integral of something like$ g(x)=0$ for $x<0$ and $g(x)=x$ for $x geq 0$
$endgroup$
– Eric
Dec 6 '18 at 17:04












1 Answer
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$begingroup$

So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
    $endgroup$
    – zhw.
    Dec 6 '18 at 17:32










  • $begingroup$
    I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
    $endgroup$
    – Andrei
    Dec 6 '18 at 19:22













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$begingroup$

So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
    $endgroup$
    – zhw.
    Dec 6 '18 at 17:32










  • $begingroup$
    I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
    $endgroup$
    – Andrei
    Dec 6 '18 at 19:22


















1












$begingroup$

So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
    $endgroup$
    – zhw.
    Dec 6 '18 at 17:32










  • $begingroup$
    I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
    $endgroup$
    – Andrei
    Dec 6 '18 at 19:22
















1












1








1





$begingroup$

So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$






share|cite|improve this answer









$endgroup$



So you need a continuous function, and the first derivative at $0$ must be the same for both $x>0$ and $x<0$, but the second derivative must be different. Start with $f(x)=x^2$ for $x>0$. You have $f(0+)=0$, $f'(0+)=0$, and $f''(0+)=2$. You now need $f(0-)=0$, $f'(0-)=0$, and you can choose $f''(0-)=0$. For example $f(x)=x^4$ for $x<0$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 16:59









AndreiAndrei

12k21126




12k21126












  • $begingroup$
    Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
    $endgroup$
    – zhw.
    Dec 6 '18 at 17:32










  • $begingroup$
    I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
    $endgroup$
    – Andrei
    Dec 6 '18 at 19:22




















  • $begingroup$
    Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
    $endgroup$
    – zhw.
    Dec 6 '18 at 17:32










  • $begingroup$
    I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
    $endgroup$
    – Andrei
    Dec 6 '18 at 19:22


















$begingroup$
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
$endgroup$
– zhw.
Dec 6 '18 at 17:32




$begingroup$
Not sure why you mention continuity. The desired function should be differentiable, hence it will be continuous. You wrote "the first derivative at $0$ must be the same for both $x>0$ and $x<0$" What does that mean?
$endgroup$
– zhw.
Dec 6 '18 at 17:32












$begingroup$
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
$endgroup$
– Andrei
Dec 6 '18 at 19:22






$begingroup$
I've started from scratch. So in order to find a differentiable function, the first requirement is that is continuous. In order to be differentiable, the derivative must be continuous, including at $0$. Since I choose to define my function differently for $x>0$ and for $x<0$, I just need $$lim_{xto 0}f(x<0)=lim_{xto 0}f(x>0)$$
$endgroup$
– Andrei
Dec 6 '18 at 19:22




















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