Question

Question #9685a

Answer 1

We don't have to look at the `(x-7)^2` as this will always be `>=0`
Of course `x!=7`

Explanation:

For `AxxB<0` one of them must be `<0` and the other `>0`

(a) `x+2<0->x<-2`
This will make `x-9<-2-9or x-9<-11`. Now both are negative. This will not work.

(b) `x-9<0->x<9`
Remember `x> -2` or we are in situation (a)

Conclusion:
`x<9andx> -2` or in short: `-2 < x<9`, but `x!=7`
graph{(x+2)(x-7)^2(x-9) [-34.98, 38.12, -31.26, 5.3]}