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Let $f: \big[0, \frac{1}{2} \big] \to \mathbb R, e^{-2x}f(x)$ is twice differentiable function having local minima at $x=\dfrac{1}{4}$ and $$\dfrac{d^2}{dx^2}\bigg(e^{-2x}f(x) \bigg) \gt0, \qquad \forall x \in \big( 0,\frac{1}{2} \big)$$ If $f(0)=f\bigg(\dfrac{1}{2}\bigg)=0$, then prove that $$\dfrac{f'\bigg(\dfrac{3}{8}\bigg)}{f\bigg(\dfrac{3}{8}\bigg)} \gt 2$$ $$\dfrac{f'\bigg(\dfrac{1}{8}\bigg)}{f\bigg(\dfrac{1}{8}\bigg)} \lt 2$$


My Attempt: Let $g(x)=e^{-2x}f(x)$.
Now according to the given conditions, I thought of the curve as a parabola with a minima at $x=\dfrac{1}{4}$ and $g(x)=0$ at $x=0,\dfrac{1}{2}$. Also, because the graph always has an upwards concavity in the given range.
$g(x)=ax^2-\dfrac{ax}{2}, a\gt 0$ but now the range does not satisfy. I also don't know which equation to apply for obtaining the respective proving conditions now. Also what can be some alternate methods? The answer key states that given condition is false and the opposite inequality of both is correct.
Thank You

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    $\begingroup$ "I thought of the curve as a parabola". This was a mistake. There is nothing in the conditions that requires $g$ to be parabolic. Because of the second derivative condition, it will look a little like a parabola, but there are plenty of non-parabolic curves with this general shape. For example concatenaries, and the family $y - y_0 = A(x - x_0)^{2n}, n > 1$. $\endgroup$ Sep 10, 2021 at 17:41
  • $\begingroup$ Okay, got your point @PaulSinclair $\endgroup$
    – UNAN
    Sep 10, 2021 at 17:44
  • $\begingroup$ @PaulSinclair can you construct a function that is not parabolic and satisfies the conditions? Your example doesn't seem to fit... $\endgroup$
    – satan 29
    Sep 10, 2021 at 17:57
  • $\begingroup$ @satan29 - I was not offering suggestions for alternative functions to try. I was pointing out that PCMSE's approach was mistaken. Instead of assuming a particular form for $g$, the proof needed to be made in general, so that it applied to all functions satisfying the conditions. $\endgroup$ Sep 10, 2021 at 18:11
  • $\begingroup$ That is indeed correct, and that's how approached my answer, but just for curiosity, can it be shown that other functions except parabolic exist, that satisfy the given conditions? $\endgroup$
    – satan 29
    Sep 11, 2021 at 6:16

2 Answers 2

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The answer key is correct in my opinion.

Let $h(x)=\dfrac{f'(x)}{f(x)} -2.$

Notice that $\dfrac{y'}{y} = \dfrac{d\ln(y)}{dx} $. It appears that we can make use of this since we have a term of $f'(x)/f(x)$ in our expression for $h(x)$.

Let $g(x) = e^{-2x}f(x)$ . Then, $f(x)= e^{2x} g(x)$ and hence $\ln(f(x))= 2x+\ln(g(x))$. Differentiating both sides, we get $ \dfrac{f'(x)}{f(x)} = 2 + \dfrac{g'(x)}{g(x)}$, which implies $h(x)=\dfrac{g'(x)}{g(x)}$.

Now, $g''(x)>0$ in $(0,1/2)$, and $x=1/4$ is a point of local minima of $g(x)$. Therefore, $g'(x)<0$ in $(0,1/4)$ and $g'(x)>0$ in $(1/4,1/2)$. Also, it's easy to see that $g(x)<0$ in $(0,1/2)$. Hence, $\dfrac{g'(x)}{g(x)}=h(x)$ is $>0$ in $(0,1/4)$, and $<0$ in $(1/4,1/8)$. Which means $h(1/8)>0$ and $h(3/8) <0$.

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  • $\begingroup$ My assumption of parabola, is it correct @satan29? Also, how did you deduce $g(x) \lt $0? $\endgroup$
    – UNAN
    Sep 10, 2021 at 11:09
  • $\begingroup$ Yes that's correct, I had discussed that with the OP after posting my answer. Thanks for the formatting tips! $\endgroup$
    – satan 29
    Sep 10, 2021 at 17:52
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    $\begingroup$ FYI - I deleted a comment that no longer matched the updated answer. But realized afterwards that it now looks like satan 29 is agreeing with PCMSE's question about being a parabola (which is false). the "Yes that correct" was addressed to my comment. $\endgroup$ Sep 10, 2021 at 18:15
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You are on the right track. $$g(x)=ax^2-\frac{ax}2=e^{-2x}f(x)$$

Differentiating with respect to $x$, we get $$\Rightarrow 2ax-\frac a2=e^{-2x}(-2f(x)+f'(x))$$

Putting $x=\frac 38$ in this expression, we get $$-2f(\frac 38)+f'(\frac 38)=\frac a4\cdot e^{\frac 34}>0$$ $$2f(\frac 38)<f'(\frac 38)$$ Now since $f(\frac 38)<0$ for $x$ in $(0,\frac 12)$, we get $$\Rightarrow \frac{f'(\frac 38)}{f(\frac 38)}<2$$ Now you can solve in a similar fashion for $x=\frac 18.$

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  • $\begingroup$ What about the range then @RiverX15, mapping to all real numbers? $\endgroup$
    – UNAN
    Sep 10, 2021 at 8:11
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    $\begingroup$ @PCMSE It's not mentioned in the question that $f$ is onto. $\endgroup$
    – RiverX15
    Sep 10, 2021 at 8:44
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    $\begingroup$ from your definition of $f(x)$ , its clear that $f(x) <0$ in $(0,1/2)$ and hence the sign of the inequality must change when you divide by $f(x)$ in the last step to obtain $f'(x)/f(x)$ $\endgroup$
    – satan 29
    Sep 10, 2021 at 9:47
  • $\begingroup$ @satan 29 Thank you for reporting the error! I did not pay much heed to the point mentioned by you and the op also accepted what I mentioned in the comment (now deleted). $\endgroup$
    – RiverX15
    Sep 10, 2021 at 14:34
  • $\begingroup$ @RiverX15, ya even I overlooked that. Actually even the textbook solution stopped before dividing. That's why I thought you got the correct one. But now yes understood. Thank you very much for your answer :-) $\endgroup$
    – UNAN
    Sep 10, 2021 at 15:02

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