Look for points where the denominator is equal to zero. In other words, solve the equation:
denominator = 0
This is a rational function; such functions have discontinuities when their DENOMINATOR (the bottom part) is equal to zero. Therefore, to find the discontinuities, simply solve the equation:Denominator = 0 Or specifically in this case: 2x + 16 = 0
They are at x = -3 and x = -2.
In such cases, there is usually a discontinuity when the denominator is zero. In other words, solve for:x + 2 = 0
Yes. A well-known example is the function defined as: f(x) = * 1, if x is rational * 0, if x is irrational Since this function has infinitely many discontinuities in any interval (it is discontinuous in any point), it doesn't fulfill the conditions for a Riemann-integrable function. Please note that this function IS Lebesgue-integrable. Its Lebesgue-integral over the interval [0, 1], or in fact over any finite interval, is zero.
Since x represents a single number, and it is x squared over x squared, then it will be the same numbers in the numerator and the denominator, no matter what value you replace x with (as long as you replace both x's with the same number). Therefore the answer is 1, unless the value of x is 0, in which case it is undefined. eg: 5 squared / 5 squared = 1 100 squared / 100 squared = 1 Try it with your calculator.
This is a rational function; such functions have discontinuities when their DENOMINATOR (the bottom part) is equal to zero. Therefore, to find the discontinuities, simply solve the equation:Denominator = 0 Or specifically in this case: 2x + 16 = 0
It is x - y + 2 = 0
They are at x = -3 and x = -2.
It is the straight line through the points (0, -1) and (1, 0).
In such cases, there is usually a discontinuity when the denominator is zero. In other words, solve for:x + 2 = 0
True!
((15xy2)/(x2+5x+6))/((5x2y)/(2x2+7x+3)) =(15xy2/5x2y)*(2x2+7x+3)/(x2+5x+6) =(3y/x)*(((2x+1)(x+3))/((x+2)(x+3) =(3y(2x+1))/(x(x+2)) =(6xy+3y)/(x2+2x)
-b + or - the square root on b squared - 4 times a times c over 2
Diverge!
Yes. A well-known example is the function defined as: f(x) = * 1, if x is rational * 0, if x is irrational Since this function has infinitely many discontinuities in any interval (it is discontinuous in any point), it doesn't fulfill the conditions for a Riemann-integrable function. Please note that this function IS Lebesgue-integrable. Its Lebesgue-integral over the interval [0, 1], or in fact over any finite interval, is zero.
One over A squared or A to the negative 2.
2