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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.

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Note that $8^x=(2^x)^3$ and $4^x=(2^x)^2$. Find the derivative. You might or might not want to let $w=2^x$. If you do, remeber that $w>0$. – André Nicolas Jan 10 '12 at 21:48
Well, actually I tried this: I found the derivative, then I found when it's not zero(I got another equation, that actually is quadratic), then I found when this equation has no roots, so that our derivative can not have zero value, but got some doubtful result. If only could you help me out a bit, it'd be very nice. – user1131662 Jan 10 '12 at 22:01
@user1131662 I changed the title to make it more accurately reflect your question, and made some minor phrasing changes. – Alex Becker Jan 10 '12 at 22:58
@user1131662: I posted the answer I had in mind earlier. Just differentiate. One has to be careful at $b=-4$. – André Nicolas Jan 11 '12 at 0:49
Thanks, everyone, it really made me happy! – user1131662 Jan 11 '12 at 3:41
up vote 1 down vote accepted

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 $b=-4$.

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Wow, great I almost got it except zero value. My 'b' is >1 instead. Thanks a lot, I'm new to this site, so excuse me for some lack of skill. – user1131662 Jan 11 '12 at 3:39
@user1131662 No problem. This site is for all kinds of math, so you're welcome. I suggest you read the faq though so that you know the rules of the site and how it works (not that you've broken any, it's just good practice). – Alex Becker Jan 11 '12 at 3:45

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.

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Thank you very much, your solution is way simplier than mine. – user1131662 Jan 11 '12 at 3:40

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