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    Maxima and Minima


    We can use differentiation to find out the maxima and minimum values of the given function. Maxima and minimum values are also known as the critical values of the given function.When the first order derivative of the function that is dy / dx = 0 and the second order derivative of the function that is d^2y / dx^2 is greater than zero then the function is said to have a minimum value.

    If the first order derivative of the function that is dy / dx = 0 and second order derivative of the function that is d^2y / dx^2 is less than zero then the function is said to have a maximum value. If the second order derivative of the function that is d^2 y / dx^2 is exactly equal to zero then the function is said to have both minimum or maximum values at one point.

    Let us discuss maxima and minima with some of the examples.

    Consider the first example be: find out the maxima and minimum values of the given function y = x^3 – 3x + 2.

    The given mathematicsfunction is y = x^3 – 3x + 2. Now first let us differentiate the given function then we will get it as dy / dx = dy / dx (x^3 – 3x + 2) which is equal to dy / dx = 3x^2 – 3. Now let us equate the first differentiation to 0 then we will get it as 3x^2 – 3 = 0 in this take 3 as common then we will get it as 3(x^2 – 1) = 0 now take 3 to the left hand side then we will get it as x^2 – 1 = 0 this is in the form of (a)^2 – (b)^2 which is equals to (a + b)(a – b), so now we will get it as (x + 1)(x – 1). Now we will get the x values as 1 and -1. Now let us substitute the given x values in the given function then we will get it as if x = 1 then y = (1)^3 – 3(1) + 2 which is equals to 1 - + 2 = 0. If x = -1 then we will get it as y = (-1)^3 – 3(-1) + 2 which is equals to -1 + 3 + 2 = 4. These are known as stationary points. Now let us find out second order derivative then we will get it as d^2y / dx^2 = 6x. in this if we substitue x = -1 then we will get it as 6(-1) = -6 which is less than 0 so at this point the function has maximum value. If we substitute x = 1 we will get it as 6(1) = 6 which is greater than 0 so at this point the function has minimum value.

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