3x^2 - 13x - 10

The function is increasing if the fnction is positive for x values:

3x^2 - 13 x -10 > 0

let us factor:

==> (3x +2)(x-5) >0

Then the function is positive is:

3x+2 > 0 and x-5 > 0

==> x > -2/3 and x >5

==> b belongs to (5, inf)

Also the function is positive is:

3x+2 < 0 and x-5 < 0

==> x < -2/3 and x< 5

==> x belongs to (-inf, -2/3)

Then the function is increasing for the intervals (-inf, -2/3) U (5,inf)

To determine the values of x for the expression 3x^2-13x-10 is positive, first we have to determine the values of x for the expression is cancelling, namely, the roots of the equation

3x^2-13x-10 = 0

So, let's apply the quadratic formula to calculate the roots:

x1 = [13+sqrt(169-120)]/6

x1 = (13+sqrt49)/6

x1 = (13+7)/6

**x1 = 10/3**

x2 = (13-7)/6

x2 = 6/6

**x2 = 1**

The constraint that the expression to be positive could be written as the inequality:

3(x - 10/3)(x-1)>0

We'll divide by 3:

(x - 10/3)(x-1)>0

Now, we'll discuss the inequality:

- the product is positive if the factors are both positive:

x - 10/3>0

x>10/3

and

x-1>0

x>1

So, x belongs to the interval (10/3 , +inf.)

- the product is positive if the factors are both negative:

x - 10/3<0

x<10/3

x-1<0

x<1

So, x belongs to the interval ( -inf.,1)

Finally, the solution set of the inequality is the union of the sets identified above: **( -inf.,1) U (10/3 , +inf.).**

**The expression is positive for the sets **

**( -inf.,1) U (10/3 , +inf.).**