Power function of fixed numbers. Prove that $3^x-4^x+2x4^{x-1}\le0$, where $x\in[-0.5,0]$.Here is it's plot.
I tried to do it by first and second derivative test but it involves $log$  which make the expression more complicated.
 A: If $y \in [0,\frac12)$, then $\left(\frac43\right)^y \le 1 + \frac12y$, because $\left(\frac43\right)^y$ is a convex function of $y$, and the inequality is true at the end-points $0$ and $\frac12$.
Now putting $x=-y$, we have, for $x \in [-\frac12,0]$,
$$\left(\frac34\right)^x \le 1 - \frac12x$$
Multiplying by $4^x$: $$3^x \le 4^x(1-\frac12x)$$
Rearranging, and using $\frac12 4^x = 2\cdot 4^{x-1}$:
$$3^x-4^x+2x4^{x-1}\le0$$
A: Using the first derivative you can see that the function is increasing on the given interval. Since $f(0) \leq 0$, the function is everywhere below 0 on the given interval. 
A: Note that
$$
f(x)=\frac x2+\left(\frac34\right)^x
$$
is a Convex Function. This
can be shown by noting that its second derivative is positive:
$$
\begin{align}
f''(x)
&=\log\left(\frac34\right)^2\left(\frac34\right)^x\\
&\ge0
\end{align}
$$
Thus, if $f(a)\le1$ and $f(b)\le1$ then for all $x$ between $a$ and $b$, $f(x)\le1$.
Since $f(-0.5)=-\frac14+\frac2{\sqrt3}\lt1$ (since $\frac43\le\frac{25}{16}$) and $f(0)=1$, we have that for all $-0.5\le x\le0$,
$$
\frac x2+\left(\frac34\right)^x\le1
$$
which means that
$$
3^x\le\left(1-\frac x2\right)4^x
$$
which implies
$$
3^x-4^x+2x4^{x-1}\le0
$$
