It depends on the function and the level of your mathematical skills. It also depends on whether you are looking for a global maximum or a local one. For example, a cubic equation [y = ax3 + bx2 + cd + d] has no global maximum but it will usually have a local one. This is also the case for any equation that is asymptotically +infinity somewhere in its domain.
If a function is twice differentiable over the domain in question, differentiate it once, set the resulting derivative equal to zero and solve for the coordinates of the stationary point. Next, differentiate it again and evaluate the value of the second derivative at the stationary point. If this derivative is negative, you have a local maximum at the stationary point. But be careful at the edges of the domain.
All this does not help if the function is not twice differentiable. Sometimes there are other ways. For example, let P(X = x) be the [probability distribution] function that X, the sum of the numbers on two dice, is x.
Then it can easily be shown that
P(2) = P(12) = 1/36
P(3) = P(11) = 2/36
P(4) = P(10) = 3/36
P(5) = P(9) = 4/36
P(6) = P(8) = 5/36
P(7) = = 6/36
and P(x) = 0 elsewhere.
P is not a continuous function and so cannot be differentiable, but the table above shows that the maximum of the function is at P(7).
Addition is the maximum or minimum function in math.
Both the function "cos x" and the function "sin x" have a maximum value of 1, and a minimum value of -1.
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.
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Two.
By taking the derivative of the function. At the maximum or minimum of a function, the derivative is zero, or doesn't exist. And end-point of the domain where the function is defined may also be a maximum or minimum.
To determine the maximum displacement, you need to calculate the peak value of the displacement function. This is done by finding the extreme values (maximum or minimum) of the function, typically by taking the derivative and setting it to zero to find critical points. Once you have these critical points, evaluate the function at those points to find the maximum displacement.
Find the maximum and minimum values that the function can take over all the values in the domain for the input. The range is the maximum minus the minimum.
Set the first derivative of the function equal to zero, and solve for the variable.
You cannot. The function f(x) = x2 + 1 has no real zeros. But it does have a minimum.
y=2x2-3x2-12x+5=0
Addition is the maximum or minimum function in math.
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.
The general procedure is to find the function's derivative, and then solve for (derivative of the function) = 0. Each of these solutions may be a local maximum or minimum - or none. Further analysis is required. A local maximum or minimum may also occur at points where the derivative is undefined, as well as at the function's endpoints (assuming it is only defined for a certain range, for example, from 0 to 10).
The minimum is the vertex which in this case is 0,0 or the origin. There isn't a maximum.....
It depends on the function whose maximum you are trying to find. If it is a well behaved function over the domain in question, you could differentiate it and set its derivative equal to 0. Solve the resulting equation for possible stationary points. Evaluate the second derivative at these points and, if that is negative, you have a maximum. If the second derivative is also 0, then you have to go to higher derivatives (if they exist). If the function is not differentiable, you may have a more difficult task at hand.
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