# Inequality related to Taylor polynomial

Consider a function defined in $$\mathbb{R}$$ such that $$f^{(3)}$$ is contiuous in $$[0,1].$$ Suppose that $$f(0) = f^{'}(0) = f^{''}(0)=f^{'}(1) =f^{''}(1)=0$$ and $$f(1)=1.$$ Prove that there exists $$c \in [0,1]$$ such that $$f^ {(3)}(c) \geq 24.$$

I tried to apply the Taylor polynomial at zero and $$1$$ . But I am getting anywhere. Someone could help me?

• How can you write the Taylor expansion when you possibly don't have all the derivatives? – YiFan Jul 18 at 23:51
• There was a typo in the question . the correct is $f^{(3)}(c) \geq 24$ . Please forgive me – math student Jul 19 at 0:34
• I’ve edited back to the very first version to match with the answers. The version before editing already has an answer here:math.stackexchange.com/questions/3297300/… – Feng Shao Jul 19 at 3:20
• People spent time to find counter examples for your first-version question, which is a good question after all. You can ask your desired question in another post but should leave the mistakes in the first version here. – Feng Shao Jul 19 at 3:23
• If the problem is $f^{(3)}(c) \geq 24$ instead of $f(c)\ge{24}$ then the question becomes much less easy but more interesting. – Piquito Jul 19 at 16:50

I'm afraid you're wrong. It is not difficult to calculate a polynomial that satisfies your conditions. Take $$f(x)=6x^5-15x^4+10x^3$$ so you have $$f'(x)=30x^4-60x^3+30x^2\\f''(x)=120x^3-180x^2+60x$$ You can verify that $$f(0) = f^{'}(0) = f^{''}(0)=f^{'}(1) =f^{''}(1)=0$$ and that $$f(1)=1$$ All your condition are satisfied for all $$c$$ in $$[0,1]$$ you have $$f(c)\lt24$$ (the maximum is almost equal to $$1$$).

• Obviously that the third derivative is continuous. I have omitted this by distraction but I have though about it when constructing the polynomial. – Piquito Jul 19 at 0:06
• There was a typo in the question . the correct is $f^{(3)}(c) \geq 24$ . Please forgive me – math student Jul 19 at 0:35
• Nothing to forgive. Those typos they usually happen. Regards. – Piquito Jul 19 at 11:18

Unfortunately wrong. $$\exists f\in C^{3}[0,1]\ \forall c \in [0,1]\colon f(c) < 24.$$

Here's such an $$f$$ as a counterexample:

Consider the polynomial $$f(x) = 6x^5 -15x^4 + 10x^3 \in C^\infty(\mathbb R)$$ \begin{aligned} f(x) &= (6x^2 -15x + 10)x^3 &f(0)&=0 &f(1)=1\\ f'(x) &= 30(x-1)^2x^2 &f'(0)&=0 &f'(1)=0\\ f''(x) &= 120(x^2-1,5+1)x &f''(0)&=0 &f''(1)=0 \end{aligned}

If we assume $$\exists c\in[0,1]\colon f(c)\ge24$$, then there must $$\exists\xi\in(0,1)$$ such that $$f(\xi)$$ is an extremum because $$f(0) < f(1) < 24$$ and $$f$$ is continuous. Since $$f$$ is continuous and differentiable: $$f'(\xi) = 0$$ must hold. But we can simply check $$f'(x)=0 \implies x\in\{0,1\} \implies x\notin(0,1) \implies \nexists\xi\in(0,1)\colon\ f'(\xi) = 0 \implies \nexists c\in[0,1]\colon f(c)\ge24.$$

• Nice! Small nitpick: your statement (in the logic symbols) is technically wrong, because it doesn't specify that $f$ satisfies the conditions of the question. – YiFan Jul 18 at 23:54
• Sorry, but I can't spot an error. Can you please explain more precisely what you think is wrong or just edit the answer? – quiliup Jul 19 at 0:22
• There was a typo in the question . the correct is $f^{(3)}(c) \geq 24$ . Please forgive me – math student Jul 19 at 0:35