For what values of $b$ does this function lack extrema? I need to find the range of values of the parameter $b$ for which the function below has no extrema. 
$$ \frac{b}{3}\,8^x   + (2) 4^x + (b+3)  2^x + b  \ln(2)$$
In the beginning I thought it'd be easy, Very simple, just finding a range when derivative is not zero. But I haven't succeeded. Could anyone please help me out with this problem? I'd be very grateful.

$\ln(x)$ is the natural logarithm.
 A: Let $f(b,x) = \frac{b}{3}8^x + 2 \cdot 4^x + (b+3) 2^x + b \log(2)$. What you want is the set of values for $b$ such that $\frac{\partial}{\partial x}f(b,x)\neq 0$ for all $x\in\mathbb{R}$. Computing the partial derivative with respect to $x$ gives
$\begin{eqnarray*}
\frac{\partial}{\partial x}f(b,x) &=& \frac{\partial}{\partial x}\left(\frac{b}{3}8^x + 2 \cdot 4^x + (b+3) 2^x + b \log(2)\right)\\
&=& \frac{b}{3}\frac{\partial}{\partial x}8^x + 2 \frac{\partial}{\partial x}4^x + (b+3) \frac{\partial}{\partial x}2^x + b \log(2)\frac{\partial}{\partial x}1\\
&=&\frac{b}{3}\ln(8)8^x+2\ln(4)4^x+(b+3)\ln(2)2^x\\
&=&\ln(2)2^x(b(4^x+1)+2^{x+2}+3)
\end{eqnarray*}$
and setting this equal to $0$ gives us $b(4^x+1)+2^{x+2}+3 = 0$, which clearly cannot happen for $b\geq 0$. However, for $-4\leq b<0$ we can set $x=-1$ to get
$$b(4^{-1}+1)+2^{-1+2}+3= \frac{5b}{4}+5\geq 0$$
while for any $b<0$ the $4^x$ term is eventually much larger than the others so for sufficiently large values of $x$,
$$b(4^x+1)+2^{x+2}+3\approx b4^x<0$$
hence we must have some value of $x$ between $-1$ and $\infty$ such that $b(4^x+1)+2^{x+2}+3 = 0$ when $-4\leq b<0$. If $b< -4$ we get
$$b(4^x+1)+2^{x+2}+3< -4(4^x+1)+2^{x+2}+3=-(2^{x+1}-1)^2\leq 0$$
and so we cannot have $b(4^x+1)+2^{x+2}+3 = 0$.
This shows that $f(b,x)$ has no extrema (treated as a function in $x$) for $b\geq 0$ or $b<-4$, and has extrema when $-4\leq b<0$.
EDIT: As Andre points out, the case $b=-4$ is special and I should have treated it differently. In this case we have only $1$ root for $-4(4^x+1)+2^{x+2}+3$, and this root is an extremum (which is otherwise not the case) so it reflects an inflection point of $f(b,x)$, rather than an extremum. So in fact there are no extrema for $b=-4$.
A: The solution below is close to Alex Becker's, though the notation used is quite different.   Let $w=2^x$, and let 
$$f(w)=\frac{b}{3}w^3+2w^2+(b+3)w.$$
Note that $w$ is constrained to be positive. Differentiate. We get
$$f'(w)=bw^2+4w+b+3.$$
If $b\ge 0$, then $f'(w)\ge 3$ for all positive $w$. So there are no (positive) extrema if $b\ge 0$. Suppose now that $b$ is negative.  We have 
$$f'(w)=(1/b) [b^2w^2+4bw+b^2+3b]=(1/b)[(bw+2)^2+b^2+3b-4].$$
Since
$$y^2+3y-4=(y+4)(y-1),$$
the polynomial $y^2+3y-4$ is $\ge 0$ if $b\le -4$. So if $b\le -4$, then $f'(w)$ does not change sign at any positive  $w$.  We conclude that if $b \ge 0$ or $b \le -4$, then $f$ cannot have an extremum at any positive $w$. 
But if $-4<b<0$, then $f$ has an extremum at a positive $w$.  This is because in this interval, $b^2+3b-4$ is negative.  So there are is a positive $w$ such that $(bw+2)^2=-(b^2+3b-4)$, that is, where $f'(w)=0$, and $f'$ changes sign.  It follows that $f$ reaches a local extremum for positive $w$. 
Note that the case $b=-4$ is special in that $f'(w)$ has a positive root. But this root is a double root, so there is "almost"  a positive extremum.    
