For which values of m does the equation: $3 \ln x+m x^3 = 17$ have $1$ solution? $2$ solutions? $0$ solution?
Thanks.
For which values of m does the equation: $3 \ln x+m x^3 = 17$ have $1$ solution? $2$ solutions? $0$ solution?
Thanks.
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 $r$, $f(r)=17$.
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.