y=2x-3 has no discontinuities because no matter what you plug into x the function will be continuous.
THEOREM 2.7.2
Polynomials are continuous functions
If P is polynomial and c is any real number then
limx → c p(x) = p(c)
2X3 2 + 2 + 2 = 6 3 + 3 = 6 ^ thats how ^
Discontinuities in mathematics refer to points on a function where there is a break in the graph. They can occur when the function is not defined at a particular point or when the function approaches different values from the left and right sides of the point. Common types of discontinuities include jump discontinuities, infinite discontinuities, and removable discontinuities.
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
1x2 1x3 1x4 and so on... 2x3 2x4
Not every function with discontinuities is integrable. A function is considered integrable (in the Riemann sense) if the set of its discontinuities has measure zero. Functions with too many or too "wild" discontinuities, such as the Dirichlet function (which is 1 for rational numbers and 0 for irrational numbers), are not Riemann integrable. However, they may still be Lebesgue integrable under certain conditions.
your equation is this... 2x3 + 11x = 6x 2x3 + 5x = 0 x(2x2 + 5) = 0 x = 0 and (5/2)i and -(5/2)i
(2x3)+(3x5)-(3x2)= 2x3=6 3x5=15 3x2=6 So..... 6x25-6= 6x25=150 150+6=156
f'(x) = 1/(2x3 + 5) rewrite f'(x) = (2X3 + 5) -1 use the chain rule d/dx (2x3 + 5) - 1 -1 * (2x3 + 5)-2 * 6x2 - 6x2(2x3 + 5) -2 ==================I would leave like this rather than rewriting this
no
21
False
What are the factors? 2x3 - 8x2 + 6x = 2x(x - 1)(x - 3).