Find the range of values of X for which a) x-2/x+3 <0 b) x^3 - 2x^2-2>0

Answer 1

a) To find the range of values of x for which x-2/x+3 < 0, we can first find the critical points where the inequality changes its sign.

Set x-2/(x+3)=0:

x-2 = 0
x = 2

Set x+3=0:

x = -3

Now we can check the intervals to see which interval satisfies the inequality:

1) For x <-3:

Substitute x=-4 into x-2/x+3:

(-4-2)/(-4+3)=-6/-1=6, which is greater than 0. Hence, x<-3 is not a solution.

2) For -3<x<2:

Substitute x=0 into x-2/x+3:

(0-2)/(0+3)=-2/3, which is less than 0. Hence, -3<x<2 is a solution.

3) For x >2:

Substitute x=3 into x-2/x+3:

(3-2)/(3+3)=1/6, which is greater than 0. Hence, x>2 is not a solution.

So, the range of values of x for which x-2/x+3 < 0 is -3<x<2.

b) To find the range of values of x for which x^3 - 2x^2 - 2 > 0, we first find the critical points where the inequality changes its sign.

Factor the polynomial:

x^3 - 2x^2 - 2 = x^2 (x - 2) - 2

We can see that x = 0 and x = 2 are critical points. Now we can check the intervals to see which interval satisfies the inequality:

1) For x<0:

Substitute x=-1 into x^3 - 2x^2 - 2:

(-1)^3 - 2(-1)^2 - 2 = -1 - 2 - 2 = -5, which is less than 0. Hence, x<0 is not a solution.

2) For 0<x<2:

Substitute x=1 into x^3 - 2x^2 - 2:

(1)^3 - 2(1)^2 - 2 = 1 - 2 - 2 = -3, which is less than 0. Hence, 0<x<2 is not a solution.

3) For x>2:

Substitute x=3 into x^3 - 2x^2 - 2:

(3)^3 - 2(3)^2 - 2 = 27 - 18 - 2 = 7, which is greater than 0. Hence, x>2 is a solution.

So, the range of values of x for which x^3 - 2x^2 - 2 > 0 is x > 2.