1.) x^{3}+27/x^{2}-3x-10 < 0

2.) (3x-1) (2-5x)/2x+1 > 0

3.) 2/x+2< or equal to x/x+1

Guest Jul 16, 2017

#1**+1 **

I'm assuming this is

1.) (x^3+27) / (x^2-3x-10) < 0 factoring the numerator and denominator, we have

[(x + 3) (x^2 - 3x + 9)] / [ ( x - 5) (x + 2) ] < 0

The second factor in the numerator is always positive so the numerator is < 0 whenever x < -3

The denominator is undefined whenever x = 5 or x = -2

And the denominator will < 0 whenever -2 < x < 5

So......the function will be < 0 whenever the numerator is positive and the denominator is negative or when the numerator is negative and the denominator is positive

The first situation will occur when -2 < x < 5

And the second wil occur when x < - 3

So the solution set is [ x < - 3 ] U [ -2 < x < 5 ]

Here's the graph :

https://www.desmos.com/calculator/xgh0b1qmvm

CPhill
Jul 16, 2017

#2**+1 **

I'm assuming that we have :

2.) (3x-1) (2-5x) / (2x+1) > 0

Setting each factor to 0 and solving for x, the critical values are :

x = 1/3 , x = 2/5 and x = -1/2

The function will be undefined when x = - 1 /2

The numerator is positive when 1/3 < x < 2/5

And the denominator is positive when x > -1/2

And the function is > 0 whenever the numerator and denominator have the same sign

So...the function is > 0 when

1/3 < x < 2/5 or when x < -1/2

Here's the graph :

https://www.desmos.com/calculator/utup8jcqp2

CPhill
Jul 16, 2017

#3**+1 **

I'm assuming that the last one is :

2 / (x+2) ≤ x / (x+1)

Setting each factor to 0 and solving for x, the critical values are -2 and - 1

Note that the inequality is undefined at these two x values

Turn this into an equality

Cross-multiplying, we have

2 (x + 1) = x ( x + 2) simplify

2x + 2 = x^2 + 2x subtract 2x from both sides

2 = x^2 rearranging, we have

x^2 - 2 = 0

And this will be true when x = - √2 or x = √2

And these are the other two critical values that we need

Note that the inequality will be true when

x < -2 , - √2 ≤ x < -1 and x ≥ √2

Here's a graph :

https://www.desmos.com/calculator/zu0vmqfgux

CPhill
Jul 16, 2017