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For which values of m does the equation: $$3 \ln x+m x^3 = 17$$ have $1$ solution? $2$ solutions? $0$ solution?


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2 Answers 2

up vote 3 down vote accepted

If $m\geq 0$, note that $f(x)=3\ln x +mx^3$ is always increasing, goes to positive infinity as $x$ does, and goes to minus infinity as $x$ decreases to zero. Also, it is continuous. This is enough information to answer the question in this case.

If $m\lt 0$, then $f'(x)=3(\frac{1}{x}+mx^2)$ has a unique positive root at $r=\sqrt[3]{-\frac{1}{m}}$. Thus $f$ is increasing on $(0,r)$ and decreasing on $(r,\infty)$. Also, $f$ goes to $-\infty$ as $x$ decreases to zero or as $x$ goes to $\infty$. The number of solutions in this case will be determined by the value of $f(r)$. The $3$ cases are $f(r)<17$, $f(r)=17$, and $f(r)>17$. To determine the condition on $m$ when each occurs, you can solve the logarithmic equation you get by plugging in $r$, $f(r)=17$.

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Thank you for the comment. can you explain me why is The number of solutions in the second case will be determined by the value of f(r)? – user6163 Feb 27 '11 at 9:44
@Nir: My first suggestion is to sketch the graph of $f$ for a few values of $m<0$. Consider what happens when $|m|$ is very small, and then what happens when it gets bigger. If you have done this and it hasn't helped, read on. If $f(r)<17$, then since $f$ is decreasing on $(r,\infty)$, then what can you say about $f(x)$ when $x\gt r$? Also, $f$ is increasing on $(0,r)$, so what does $f(r)<17$ imply about $f(x)$ if $0<x<r$? Similar considerations apply if $f(x)=17$. Now if $f(x)>17$ it's a little different; then you need to use what happens as $x\to0$ and $x\to\infty$, and continuity. – Jonas Meyer Feb 27 '11 at 10:01

Can't improve on Jonas Meyer's answer, but playing with this geogebra applet might help you get a feel for the equation. Move the slider to change $m$ in increments of 1/1000000.

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That's nice. However, something more like 1/10^9 is needed to show the qualitative change that occurs for negative m as |m| increases. – Jonas Meyer Feb 28 '11 at 0:40

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