- #1

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**Serius problem!!**

Can you help me in solving this??

Is there a positive and twice differentiable function f defned on [0,∞) such

that f(x)f’’(x) ≤ -1 on [0,∞) ? Why or why not???

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- Thread starter vip89
- Start date

- #1

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Can you help me in solving this??

Is there a positive and twice differentiable function f defned on [0,∞) such

that f(x)f’’(x) ≤ -1 on [0,∞) ? Why or why not???

- #2

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Guys,no one can solve this problem??!!

- #3

HallsofIvy

Science Advisor

Homework Helper

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What does that tell you about f"(x)? Where is f concave up or down?

- #4

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Guys,no one can solve this problem??!!

Is anybody supposed to solve it.........besides you????

- #5

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Any positive constant funtion works.Can you help me in solving this??

Is there a positive and twice differentiable function f defned on [0,∞) such

that f(x)f’’(x) ≤ -1 on [0,∞) ? Why or why not???

- #6

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- 4

Any positive constant funtion works.

What do you mean with constant?

say

f(x)=b, where b is a constant, this defenitely doesn't work, since f'=f''=0 so also

f(x)f''(x)=0, so this defenitely is not smaller than -1.

- #7

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SInce f(x)>0 on[0,infty) this means that the sign of f(x)f''(x) is determined only by the sign of f''. So, since the problem requires that f(x)f''(x) be smaller or equal to -1 on the interval [0,infty) it means that for all x's on this interval f''(x)<0. What does this tell us about the concavity of f? this means that f is concave down on the whole interval. But if f is concave down on the whole interval it means that at some point it should defenitely cross the x-axis, and therefore at some point be also negative, but this contradicts the fact that f>0, so i would say that such a function does not exist at all.

- #8

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Sorry I reversed the inequality. :-)What do you mean with constant?

say

f(x)=b, where b is a constant, this defenitely doesn't work, since f'=f''=0 so also

f(x)f''(x)=0, so this defenitely is not smaller than -1.

- #9

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- #10

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that is wt I did,pls send me ur help

This is not what

Anyways, could you prove that a function on [0,inf) which is always concave down has to be negative somewhere..? This is the key argument in your reasoning and you did not justify it.

- #11

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The answer is NO:

Suppose the statement is true. Then [tex]f(x)>0, f''(x)<0 [/tex] for all [tex] x>0 [/tex]. Moreover, [tex]\lim_{x\rightarrow +\infty} f''(x)<0 [/tex] if it exists.

Let [tex] f(0)>0, f'(x) [/tex] is monotonically decreasing, then [tex]\lim_{x\rightarrow +\infty} f'(x) [/tex] is either [tex] -\infty [/tex] or a constant.

If [tex]\lim_{x\rightarrow +\infty} f'(x)=-\infty [/tex], then [tex]f [/tex] is monotonically decreasing on [tex] [T,\,+\infty)[/tex] for some [tex]T [/tex] large enough and [tex]\lim_{x\rightarrow+\infty} f(x) [/tex] must be negative. Otherwise, [tex]\lim_{x\rightarrow+\infty} f(x)=c_0\geq 0 [/tex] and there exists a unbounded sequence [tex]\{x_n\}[/tex] so that [tex]\lim_{n\rightarrow +\infty} f'(x_n)=0 [/tex]. Contradiction;

if [tex]\lim_{x\rightarrow +\infty} f'(x)=c [/tex], where [tex]c[/tex] is a constant, then there exists a unbounded sequence [tex]\{x_n\}[/tex] so that [tex]\lim_{n\rightarrow +\infty} f''(x_n)=0 [/tex], but we already have [tex]\lim_{n\rightarrow +\infty} f''(x_n)<0 [/tex]. Contradiction.

P. S.

Note that the assumption is [tex]f(x)\cdot f''(x)\leq -1 [/tex] on [tex][0,\,+\infty)[/tex]. If we change it into [tex]f(x)\cdot f''(x)<0 [/tex], then we cannot have [tex]\lim_{x\rightarrow +\infty} f''(x)<0 [/tex] and in this case the statement is true:

e.g.: Let [tex]f(x)=\ln (x+2)[/tex]. Then we have

[tex]\begin{align*}

f'(x)&=1/(x+2),\\

f''(x)&=-1/(x+2)^2.

\end{align*}[/tex]

Therefore, for all [tex]x\geq 0[/tex], we havef'(x)&=1/(x+2),\\

f''(x)&=-1/(x+2)^2.

\end{align*}[/tex]

[tex]\begin{align*}f(x)\cdot f''(x)&=-\frac{\ln (x+2)}{(2+x)^2}<0 \\

\intertext{and}

f(x)&>0.

\end{align*}[/tex]

\intertext{and}

f(x)&>0.

\end{align*}[/tex]

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