Question

Guys don't skip this question please!!!!!!!!!!!!!!!

for which value of x will
1.g(x)>f(x)
2.f(x).g(x)<0​

Answer 1

I highly suspect you have forgotten to give a concrete value for
`m`.

1.
`x < (-m - √(m^2 + 72))/(6) \vee x > (-m + √(m^2 + 72))/(6)`

2.
`\left(m\leq -3\land \left(-2<x<(6)/(m)\lor x>2\right)\right)\\\lor \left(\left(-3<m<0\land \left((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<(6)/(m)\right)\right)\right)\\\vee \left(m>3\land \left(x<-2\lor (6)/(m)<x<2\right)\right)`

Explanation:

1. Solve for
`g(x) > f(x)`:

`mx + 6 > -3x^2 + 12\\3x^2 + mx - 6 > 0\\x < (-m - √(m^2 + 72))/(6) \vee x > (-m + √(m^2 + 72))/(6)`

2. Solve for
`f(x) \cdot g(x) < 0`: (solving takes a lot of time so I think you forgot to give an actual value for
`m`, but I will just give the end result here)

`(-3x^2 + 12)(mx + 6) > 0\\\left(m\leq -3\land \left(-2<x<(6)/(m)\lor x>2\right)\right)\\\lor \left(\left(-3<m<0\land \left((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<(6)/(m)\right)\right)\right)\\\vee \left(m>3\land \left(x<-2\lor (6)/(m)<x<2\right)\right)`