find all discontinuities (16x/2x3-8x3=0)
You are probably not getting a reply because this is a confusing question.
you write it as an equation, equations have solutions, we don't speak of discontinuities for equations
Secondly, is the denominator 2x^3 - 8x^3
or only 2x^3 , the way you typed it.
Thirdly, if the denominator is 2x^3 - 8x^3, why was it not simplified to -6x^3
Is there a typo??
To find the discontinuities of the equation (16x/2x^3 - 8x^3 = 0), we need to determine the values of x that cause the equation to be undefined.
First, let's simplify the equation by factoring out the common term from the numerator:
16x / (2x^3 - 8x^3) = 0
16x / (-6x^3) = 0
Now, let's identify the values of x that make the denominator zero because dividing by zero is undefined. To do this, we set the denominator equal to zero and solve for x:
-6x^3 = 0
Dividing both sides by -6:
x^3 = 0
Taking the cube root of both sides:
x = 0
Therefore, x = 0 is the only value that makes the denominator zero, resulting in a discontinuity in the equation.