Find the domain of 2x^3-5 divided by x^2+x-6
Values of x where x^2+x-6 differently of zero.
x^2+x-6=0 for
x= -3
and
x= 2
Domain: (- infinity,-3]U[-3,2]U[2,infinity)
since x^2 + x - 6 = (x+3)(x-2)
the functions has vertical asymptotes at x=-3 and x=2 , because we would be dividing by zero
domain: any real values of x , x≠-3,2
The intervals need to be open on both ends, since the endpoints are not included
To find the domain of the function (2x^3 - 5) / (x^2 + x - 6), we need to determine the values of x for which the function is defined. The function will be defined for all values of x except the ones that lead to division by zero and any other values that make the function undefined.
First, let's find the values of x that make the denominator equal to zero, which would result in division by zero. The denominator is the quadratic expression x^2 + x - 6. We can factorize it by breaking the middle term:
x^2 + x - 6 = (x - 2)(x + 3)
Setting each factor equal to zero, we find:
x - 2 = 0 or x + 3 = 0
Solving each equation, we have:
x = 2 or x = -3
These two values, x = 2 and x = -3, will make the denominator zero, so they are not in the domain of the function.
Therefore, the domain of the function (2x^3 - 5) / (x^2 + x - 6) is all real numbers except x = 2 and x = -3. In interval notation, the domain can be written as (-∞, -3) U (-3, 2) U (2, +∞).