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Let $f$ be a twice differentiable function satisfying: $$ \cases{f(1) = 1 \\ f(2) =4\\ f(3) = 9 }$$ then :

  1. $ f''(x)= 2 $ for all $ x \in R $

  2. $f'(x) = 5 = f''(x) $ for some $ x\in (1,3)$

  3. There exists at least one $ x \in (1,3)$ such that $ f''(x)=2$

Then which of the following are true?

So what I basically tried is that I assumed that the function could be $ x^2$ But the answer I got from that was wrong.

measure_theory
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Tejus
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  • Have you tried the Mean Value Theorem? – ajotatxe Sep 20 '16 at 16:23
  • @ajotatxe No but what's the harm in assuming a function ? Also could you help with how to apply Mean value theorem in this ques – Tejus Sep 20 '16 at 16:28
  • By assuming a function, you can only show that 1, 2 or 3 is not true, but you cannot prove that 1,2 or 3 is true. – Hetebrij Sep 20 '16 at 16:29
  • @Hetebrij But by taking $x^2$ as the function , 1 is coming out to be true – Tejus Sep 20 '16 at 16:32
  • By taking $f(x)=x^2$ you have found a particular twice differentiable function with the additional property that $f''(x)=2$ for all $x$. However, as Ross Millikan have shown, we can find a twice differentiable function $g$ such that $g''(x) \neq 2$ for some $x$. – Hetebrij Sep 20 '16 at 16:54

2 Answers2

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It is true that $f(x)=x^2$ satisfies the requirements, but many other functions do as well. You can imagine putting squiggles on top of $x^2$ as long as you go through the points. An example is below. enter image description here All the questions ask if they are true for all functions that satisfy the requirements, not just for one of them. My example shows that (1) can be false. You should be able to modify it to find counterexamples to the second. You do have $f'(x)=5$ for some $x$ because of the mean value theorem, but you might not have $f''(x)=5$ at that point.

Community
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Ross Millikan
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  • But the answer given is 3 . Also can you help me with applying Mean value theorem in this question. – Tejus Sep 20 '16 at 16:40
  • For the mean value theorem, read about it and look at the values of $f(2)$ and $f(3)$. You have all the requirements because $f'(x)$ is continuously differentiable. For 3, you could have an $f$ that is constant $1$ from $1$ to $3/2$, then is $1+12(x-3/2)^2$ from $3/2$ to some value above $2$, then patch something on to go through $f(3)=9$ – Ross Millikan Sep 20 '16 at 16:47
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We show that 3) is true.

Let $h(x)=f(x)-x^2$ then $h(1)=h(2)=h(3)=0$. Hence by the Mean Value Theorem,

i) there is $x_1\in (1,2)$ such that $h'(x_1)=0$;

ii) there is $x_2\in (2,3)$ such that $h'(x_2)=0$;

iii) there is $x_3\in (x_1,x_2)\subseteq (1,3)$ such that $h''(x_3)=0$ (here the MVT is applied to $f'$).

It follows that $f''(x_3)=h''(3)+(x^2)''_{x=x_3}=0+2=2$.

A counterexample for 1) and 2) is $f(x)=x^2+\frac{1}{6}(x-1)(x-2)(x-3)$. Then $$f'(x)=\frac{3x^2+11}{6} \quad\mbox{and}\quad f''(x)=x.$$ Hence $f''(x)=x=5\not\in (1,3)$.

Robert Z
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  • @Ross Millikan Where am I assuming that $f''$ is continuous? Moreover in your counterexample $f'$ is continuous and $\lim_{x\to 3/2^+}f''(x)=24$ implies that $f''(3/2)=24$ which is not (http://math.stackexchange.com/questions/530577/if-a-derivative-of-a-continuous-function-has-a-limit-must-it-agree-with-that-li?rq=1). So f is not twice differentiable at $3/2$. Am I missing something? – Robert Z Jan 25 '17 at 19:16