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CALCULUS Differentiation Integeration Today we are going to learn a particular topic in differentiation.i.e How to find Maximum and minimum value of a function.Let y = f(x) be a function. Now let us know the steps involving to find Maximum and Minimum Value of the given function.Step 1 Find f ’(x) (First Derivative) Step 2 Find f “(x) (Second derivative) dy dx d^2y dx^2Step 3 Find 0 and find x = a,b Step 4 sub x = a , x = b in the second derivation That is dy dx d^2y dx^2Case (i) At x= a , If d^2y dx^2 The Function is Minimum at x=a The Minimum Value y = f(a)Case (i) At x= b , If d^2y dx^2 The Function is Maximum at x=b The Maximum Value y = f(b)Case (iii) if when x=a,x=b we can say There is no Maximum and Minimum value for the function at x = a and x= b.Let us see the above procedure for finding the Maximum and Minimum Value of a function in the following examplesd^2y dx^2Ex :1 y = 4x^3 + 2x^2 Differentiate with respect to ‘x’ 4*3x^(3-1) + 2 * 2x^(2-1) 12x^2 + 4x dy dx dy dxAgain Differentiate with respect to x 12*2x^(2-1) + 4*1 24x+4 d^2y dx^2 d^2y dx^2Now equalize dy dx 0 12 x^2+4x=0 4x*3x + 4x = 0 4x ( 3x+1) = 0 Either 4x = 0 or 3x+1= 0 4x 0 3x + 1 = 0 4 4 -1 -1 Therefore x = 0 3x = -1 4x 0 3x + 1 = 0 4 4 -1 -1 3x = -1 x = 0 3x -1 3 3 x = -1 3When we equalize dy/dx to 0, we get two values for ‘x’ They are x = 0 and x = -1 3 Now we have to substitue this values in d^2y dx^2At x = 0 d^2y dx^2 d^2y dx^2 d^2y dx^2 = 24(0) +4 = 0 + 4 = 4 > 0Therefore at x = 0, the given function is Minimum. The Minimum value y = 4(0) +2(0) y = 0 + 0 y = 0 At x = -1 d^2y 3 dx^2 d^2y dx^2 d^2y dx^2 = 24 (-1/3) +4 = -8+4 = -4 < 0There for the given function is Maximum at x = -1/3. The Maximum value y = 4 (-1/3)^3 + 2 (-1/3)^2 y = 4(-1/27) + 2(1/9) y = -4 + 2 -4 + 6 27 9 27 27 y = -4 + 6 2 27 27