How can I find the maximum of $2\cdot 3^{-x}$? I know its close to $1$ because I have seen its graph, but when I differentiate the function and set it equal to zero (to get a maximum) I get $-2\cdot 3^{-x}=0$. What does that mean? How can I solve for the $x$ that gives the maximum $y$?
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I am very sorry, I had a typo in the function. I graphed it as y=2X3^-X... and the shape is very different. Sorry for asking such a silly question! – Sophie Oct 15 '13 at 04:01
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The derivative with respect to $x$ of $f(x)=2\cdot3^{-x}$ is
$$f\,'(x)=2\cdot 3^{-x}(-1)\ln 3=-2\cdot 3^{-x}\ln 3\;.$$
Since $\ln 3>0$ and $3^{-x}>0$ for all $x$, $f\,'(x)<0$ for all $x$, and indeed the function is decreasing everywhere and has no maximum. In fact $\lim\limits_{x\to-\infty}f(x)=\infty$.
Brian M. Scott
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The derivative of $2 \cdot 3^{-x}$ is $-2 \log(3) 3^{-x}$ It is never zero. This reflects the fact that $2 \cdot 3^{-x}$ has no maximum, but goes to $+\infty$ as $x \to -\infty$
Ross Millikan
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