+3 and -3
The only variable on the right hand side is sin(x). The maximum value of sin(x) is 1. So, the max value of 3sin(x) is 3*1 = 3 and so, the max value of 3sin(x) + 2 is 3+2 = 5.
Domain and range are not sufficient to determine the y intercept. For example, the domain and range for the straight line y = 2x + 3 are the whole of the real numbers. That tells you nothing about the intercept.
It is y >= 5.
The domain is (-infinity, infinity) The range is (-3, infinity) and the asymptote is y = -3
+3 and -3
8
The sine function (sin x) can only have values in the range between 1 and -1. Perhaps you can work it out from there.
Maybe; the range of the original function is given, correct? If so, then calculate the range of the inverse function by using the original functions range in the original function. Those calculated extreme values are the range of the inverse function. Suppose: f(x) = x^3, with range of -3 to +3. f(-3) = -27 f(3) = 27. Let the inverse function of f(x) = g(y); therefore g(y) = y^(1/3). The range of f(y) is -27 to 27. If true, then f(x) = f(g(y)) = f(y^(1/3)) = (y^(1/3))^3 = y g(y) = g(f(x)) = g(x^3) = (x^3)^3 = x Try by substituting the ranges into the equations, if the proofs hold, then the answer is true for the function and the range that you are testing. Sometimes, however, it can be false. Look at a transcendental function.
The only variable on the right hand side is sin(x). The maximum value of sin(x) is 1. So, the max value of 3sin(x) is 3*1 = 3 and so, the max value of 3sin(x) + 2 is 3+2 = 5.
Domain and range are not sufficient to determine the y intercept. For example, the domain and range for the straight line y = 2x + 3 are the whole of the real numbers. That tells you nothing about the intercept.
y=-3x*sinx-1.5x2+5x, when x=πy'=d/dx(-3x*sinx)-d/dx(1.5x2)+d/dx(5x)y'=(-3x*d/dx(sinx)+sinx*d/dx(-3x))-d/dx(1.5x2)+d/dx(5x)y'=(-3x*cosx+sinx(-3))-d/dx(1.5x2)+d/dx(5x)y'=(-3x*cosx-3sinx)-3x+5y'=-3x*cosx-3sinx-3x+5 is the derivative at any point of that equation, now you only have to plug in π for xy'(π)=-3π*cosπ-3sinπ-3π+5y'(π)=-3π*(-1)-3(0)-3π+5y'(π)=3π-3π+5y'(π)=5
Y = x squared -4x plus 3 is an equation of a function. It is neither called a domain nor a range.
It is y >= 5.
The range is 2.
The domain is (-infinity, infinity) The range is (-3, infinity) and the asymptote is y = -3
The range is (3/4 to 3/0.5) or 0.75 to 6