On the y-axis, normally.
The maximum value of X minus the minimum value of X is calculated by subtracting the minimum value from the maximum value. This difference represents the range of the values of X. If you have specific values for X, you can determine the maximum and minimum values and then compute this difference accordingly.
Suppose the revenue equation is of the form R = ax2 + bx + c where a, b and c are constants and x is the variable. To have a maximum, either a must be negative or x must lie within fixed limits. If a is negative then the maximum revenue is attained when x = -b/(2a). That is, find the value of R when x = -b/(2a). If a is positive, then find the value of R when x is at each end point of its domain. One of them will be larger and that is the maximum value of the revenue.
If x is the unknown or variable in an equation it can have many possible maximum or minimum values
To find the maximum value of (6x + 5y) in the feasible region, we typically need the constraints that define that region (such as inequalities involving (x) and (y)). Without those constraints, we can't determine the maximum. Generally, the maximum occurs at one of the vertices of the feasible region formed by the intersection of the constraint lines. If you provide the specific constraints, I can help find the maximum value.
Take the derivative of the function.By plugging a value into the derivative, you can find the instantaneous velocity.By setting the derivative equal to zero and solving, you can find the maximums and/or minimums.Example:Find the instantaneous velocity at x = 3 and find the maximum height.f(x) = -x2 + 4f'(x) = -2xf'(3) = -2*3 = -6So the instantaneous velocity is -6.0 = -2x0 = xSo the maximum height occurs at x = 0f(0) = -02 + 4 = 4So the maximum height is 4.
Assuming the standard x and y axes, the range is the maximum value of y minus minimum value of y; and the domain is the maximum value of x minus minimum value of x.
In Calculus, to find the maximum and minimum value, you first take the derivative of the function then find the zeroes or the roots of it. Once you have the roots, you can just simply plug in the x value to the original function where y is the maximum or minimum value. To know if its a maximum or minimum value, simply do your number line to check. the x and y are now your max/min points/ coordinates.
Suppose the revenue equation is of the form R = ax2 + bx + c where a, b and c are constants and x is the variable. To have a maximum, either a must be negative or x must lie within fixed limits. If a is negative then the maximum revenue is attained when x = -b/(2a). That is, find the value of R when x = -b/(2a). If a is positive, then find the value of R when x is at each end point of its domain. One of them will be larger and that is the maximum value of the revenue.
up and down. the x goes left and right
if you have any doubts ask
Both the function "cos x" and the function "sin x" have a maximum value of 1, and a minimum value of -1.
If x is the unknown or variable in an equation it can have many possible maximum or minimum values
Sin(x) has a maximum value of +1 and a minimum value of -1.
Take the derivative of the function.By plugging a value into the derivative, you can find the instantaneous velocity.By setting the derivative equal to zero and solving, you can find the maximums and/or minimums.Example:Find the instantaneous velocity at x = 3 and find the maximum height.f(x) = -x2 + 4f'(x) = -2xf'(3) = -2*3 = -6So the instantaneous velocity is -6.0 = -2x0 = xSo the maximum height occurs at x = 0f(0) = -02 + 4 = 4So the maximum height is 4.
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In theory you can go down the differentiation route but because it is a quadratic, there is a simpler solution. The general form of a quadratic equation is y = ax2 + bx + c If a > 0 then the quadratic has a minimum If a < 0 then the quadratic has a maximum [and if a = 0 it is not a quadratic!] The maximum or minimum is attained when x = -b/2a and you evaluate y = ax2 + bx + c at this value of x to find the maximum or minimum value of the quadratic.
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.