Solving $\dfrac{(3^x-4^x)\cdot \ln (x+2)}{x^2-3x-4}\leq 0$ 
The solution set of the inequality $\dfrac{(3^x-4^x)\cdot \ln (x+2)}{x^2-3x-4}\leq 0$ is.

My attempts:
For $x>0\implies 3^x-4^x<0$
$\implies \dfrac{\ln (x+2)}{x^2-3x-4}\geq 0$. Now root of numerator is $x=-1$ and of denominator is $x=4,-1$
By wavy curve method:
$\implies x\epsilon(4,\infty)$
Now for $x\leq 0\implies 3^x-4^x\geq 0$
Hence:
$\dfrac{\ln (x+2)}{x^2-3x-4}\leq 0$
$\implies x\epsilon(-1,0]$
My answer $(-1,0]\cup (4,\infty)$
But book's answer said I missed one interval, please help.
I observed something that $-1$ are common roots in both numerator and denominator, so should we put $-ve$ on LHS of $-1$? If we do this then my answer will be correct, but don't know whether it's correct thinking or not, if it is then why?
 A: HINT: since we have $$x^2-3x+4\geq 0$$ for all real $x$ it must be
$$(3^x-4^x)\ln(x+2)\le 0$$ solving this we get
$$-2<x<=-1$$ or $$x\geq 0$$
After the correction we have two cases:
$$x^2-3x-4>0$$ and $$(3^x-4^x)\ln(x+2)\le 0$$
solving this we obtain:
$$x<-1$$ or $$x>4$$ and $$-2\le x\le -1$$ or $$x>4$$
in the second case we get
$$x^2-3x-4<0$$ and $$(3^x-4^x)\ln(x+2)\geq 0$$
solving this we obatain:
$$-1<x<4$$ and $$-1\le x\le 0$$
Cand you finish?
A: Remember that the sign of the fraction is the product of the signs of the single factors, so you have to solve separately the inequalities:
$$
(x-4)(x+1) >0
$$
$$3^x-4^x\ge 0$$
and 
$$
\ln (x+2)\ge 0
$$
That can be solved in $\mathbb{R}$ with the condition  $x>-2$
Putting all solutions together we have this picture ( where in red are representd the ''negative'' intervals for the factor at right)

from which we see that the fraction is $\le 0$ in the intervals:
$$
(-2,-1)\cup(-1,0] \cup (4,+\infty)
$$
A: The term $3^x-4^x$ is positive for $x<0$ and negative for $x>0$.
The term $\ln(x+2)$ is positive for $x>-1$ and negative for $-2<x<-1$ (undefined for $x\le-2$).
The term $x^2-3x-4$ is positive for $x<-1$ or $x>4$ and negative for $-1<x<4$.
$$
\begin{array}{l|ccccccccc}
&& -2 & & -1 && 0 && 4 \\
\hline
3^x-4^x & <0 &<0 & <0 & <0 & <0 & =0 & >0 & >0 & >0 \\
\ln(x+2) & \text{u} & \text{u} & <0 & =0 & >0 & >0 & >0 & >0 & >0 \\
x^2-3x-4 & >0 & >0 & >0 & =0 & <0 & <0 & <0 & =0 & >0 \\
\hline
f(x) & \text{u} & \text{u} & >0 & \text{u} & >0 & =0 & <0 & \text{u} & >0
\end{array}
$$
where “u” means “undefined”.
The last row is obtained by using the “rule of signs” on the factors. The function is undefined where the denominator vanishes.
Hence you get, for the solutions, $(-2,-1)\cup(-1,0]\cup(4,\infty)$.
The problem with your solution is that two factors vanish at $-1$, so the global sign doesn't change when we cross it.
A: Your last comment is correct.  I'm not sure what the "wavy curve" method is, but it boils down to finding all the places where the graph of the function can cross the $x$-axis.  (Which would be where either the denominator or numerator equals $0$.)  Your function is undefined at zero, but it's limit is $-1/60$, which is not zero. The graph doesn't cross the $x$-axis at $-1$.  
A: For $x\in (-2,-1)$ we get
\begin{align*}
3^x-4^x&>0\\
\ln(x+2)&<0\\
x^2-3x-4=\left(x+1\right)\left(x-4\right)&>0
\end{align*}
Then
$$x\in (-2,-1)\quad\implies \quad \frac{(3^x-4^2)\ln(x+2)}{x^2-3x-4}<0$$
A: You missed in fact one inequality: note that
$ln(x+2)\le 0$ for $0<x+2 \le 1$, i.e. for $-2<x \le -1$
Adding this, you can complete the calculations as you were correctly doing.
