prove (3x^2 + 4x-1)/(7x^5+5)>0 is field.
can you just prove one of the addition axioms?
To prove that the inequality (3x^2 + 4x - 1)/(7x^5 + 5) > 0 is true, we need to determine specific values of x for which the expression is greater than zero.
To approach this problem, we can break it down into steps:
Step 1: Determine the critical points
Critical points are the values of x for which the expression is equal to zero or undefined. In this case, we need to find the values of x that make the numerator or denominator equal to zero.
Setting the numerator equal to zero:
3x^2 + 4x - 1 = 0
To solve this quadratic equation, you can use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac))/2a
In our case, a = 3, b = 4, and c = -1.
Plugging these values into the quadratic formula:
x = (-(4) ± sqrt((4)^2 - 4(3)(-1)))/(2(3))
Simplifying further:
x = (-4 ± sqrt(16 + 12))/6
x = (-4 ± sqrt(28))/6
So the critical points for the numerator are:
x = (-4 + sqrt(28))/6 and x = (-4 - sqrt(28))/6
Next, we need to consider the denominator. We see that 7x^5 + 5 is always positive, meaning it does not have any critical points.
Step 2: Determine the sign of the expression
We need to determine the sign of the expression (3x^2 + 4x - 1)/(7x^5 + 5) in the intervals between the critical points.
For the numerator:
When x < (-4 - sqrt(28))/6:
Substituting a value less than (-4 - sqrt(28))/6 into the quadratic equation, we get a positive value for the numerator because x^2 is positive.
When (-4 - sqrt(28))/6 < x < (-4 + sqrt(28))/6:
Substituting a value between the two critical points into the quadratic equation, we get a negative value for the numerator. This is because the parabola opens upwards, and between these two points, the value of the function is negative.
When x > (-4 + sqrt(28))/6:
Substituting a value greater than (-4 + sqrt(28))/6 into the quadratic equation, we get a positive value since the quadratic function is open upwards.
Next, we consider the denominator. As mentioned before, 7x^5 + 5 is always positive for any real value of x.
Step 3: Combine the numerator and denominator
Since the numerator is positive when x < (-4 - sqrt(28))/6 and x > (-4 + sqrt(28))/6, and the denominator is positive for all values of x, the expression (3x^2 + 4x - 1)/(7x^5 + 5) will be positive when:
x < (-4 - sqrt(28))/6 or x > (-4 + sqrt(28))/6
Thus, we have determined the values of x for which the inequality (3x^2 + 4x - 1)/(7x^5 + 5) > 0 holds.