How to obtain the relation between the function and its derivative? 
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
 A: 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 $h(3/8) <0$.
A: 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 $x=\frac 18.$
