Answer:
I highly suspect you have forgotten to give a concrete value for [tex]m[/tex].
1. [tex]x < \frac{-m - \sqrt{m^2 + 72}}{6} \vee x > \frac{-m + \sqrt{m^2 + 72}}{6}[/tex]
2. [tex]\left(m\leq -3\land \left(-2<x<\frac{6}{m}\lor x>2\right)\right)\\\lor \left(\left(-3<m<0\land \left(\frac{6}{m}<x<-2\lor x>2\right)\right)\right)\\\vee\left((m=0\land (x<-2\lor x>2))\lor \left(0<m\leq 3\land \left(x<-2\lor 2<x<\frac{6}{m}\right)\right)\right)\\\vee \left(m>3\land \left(x<-2\lor \frac{6}{m}<x<2\right)\right)[/tex]
Step-by-step explanation:
1. Solve for [tex]g(x) > f(x)[/tex]:
[tex]mx + 6 > -3x^2 + 12\\3x^2 + mx - 6 > 0\\x < \frac{-m - \sqrt{m^2 + 72}}{6} \vee x > \frac{-m + \sqrt{m^2 + 72}}{6}[/tex]
2. Solve for [tex]f(x) \cdot g(x) < 0[/tex]: (solving takes a lot of time so I think you forgot to give an actual value for [tex]m[/tex], but I will just give the end result here)
[tex](-3x^2 + 12)(mx + 6) > 0\\\left(m\leq -3\land \left(-2<x<\frac{6}{m}\lor x>2\right)\right)\\\lor \left(\left(-3<m<0\land \left(\frac{6}{m}<x<-2\lor x>2\right)\right)\right)\\\vee\left((m=0\land (x<-2\lor x>2))\lor \left(0<m\leq 3\land \left(x<-2\lor 2<x<\frac{6}{m}\right)\right)\right)\\\vee \left(m>3\land \left(x<-2\lor \frac{6}{m}<x<2\right)\right)[/tex]
