x^2/2
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
x isn't in the calculation so it cant be solved.
1/5y2+9
Multiply the coefficient by all the values inside the brackets.Example 1:2(y+1)= 2 X y + 2 X 1= 2y + 2Example 2:3 X 5y(y+4)= 3 X (5y X y) + (5y X 4)= 3 X 5y2 + 20y
The "root" of a function is also called the "zero" of a function. This is where the function equals zero. The function y=4-x2 has roots at x=2 and x=-2 The function y=4-x2 has zeroes at x=2 and x=-2 Those are equivalent statements.
False. It is true for a function that is continuous at x=2, but it is not generally true for all functions. For a counterexample, consider the function f(x), such that: f(x)=x for x not equal to 2 f(x)=0 for x=2 The limit of this function as x approaches 2 is 2 (since we can make f(x) as close to 2 as we want as x gets closer to 2), but f(2) does not equal the limit of f(x) as x approaches 2.
The function given is (f(x) = -x^2). The second derivative of a function, denoted as (f’'(x)), measures the concavity of the function. For the function (f(x) = -x^2), the first derivative (f’(x)) is (-2x). Taking the derivative of (f’(x)) gives us the second derivative (f’‘(x)), which is (-2). So, (f’'(x) = -2). This indicates that the function (f(x) = -x^2) is concave down for all (x), because the second derivative is negative.
How would you prove algebraically that the function: f(x)= |x-2|, x<= 2 , is one to one?
x^2/2
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]
A function of x means that the answers depends on the value of x that is substituted into the function. For example: y = x + 1 If x = 1, y = 2 If x = 2, y = 3 and so on.
The domain of the function f (x) = square root of (x - 2) plus 4 is Domain [2, ∞)
x2 + 8xy + 15y2 = (x + 3y) (x + 5y)