x5y2 - 2 = 0
... makes sense
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The integral of the absolute value function, |x|, is given by the piecewise function ∫|x|dx = (x^2)/2 + C for x ≥ 0 and ∫|x|dx = (-x^2)/2 + C for x < 0, where C is the constant of integration. This is because the absolute value function changes its behavior at x = 0, resulting in two different expressions for the integral depending on the sign of x.
f(x) = 2 cos 3x The amplitude: A = |2| = 2 The maximum value of the function: 2 The minimum value of the function: -2 The range: [-2, 2]
The domain of the function f (x) = square root of (x - 2) plus 4 is Domain [2, ∞)
The Equation of a Rational Function has the Form,... f(x) = g(x)/h(x) where h(x) is not equal to zero. We will use a given Rational Function as an Example to graph showing the Vertical and Horizontal Asymptotes, and also the Hole in the Graph of that Function, if they exist. Let the Rational Function be,... f(x) = (x-2)/(x² - 5x + 6). f(x) = (x-2)/[(x-2)(x-3)]. Now if the Denominator (x-2)(x-3) = 0, then the Rational function will be Undefined, that is, the case of Division by Zero (0). So, in the Rational Function f(x) = (x-2)/[(x-2)(x-3)], we see that at x=2 or x=3, the Denominator is equal to Zero (0). But at x=3, we notice that the Numerator is equal to ( 1 ), that is, f(3) = 1/0, hence a Vertical Asymptote at x = 3. But at x=2, we have f(2) = 0/0, 'meaningless'. There is a Hole in the Graph at x = 2.
Any multiple of X^2+X/2-1/2