Relative maximum point

The above graph of the function slopes upwards to the right between points C and A and thus has a positive slope among these two points. The function has a negative slope between points E and C. So at point C, the slope of the function is Zero.

Among points X1 and X2    ((dy)/(dx)) > 0; whereas X1 ≤ X < X2

And among X2 and X3   ((dy)/(dx)) < 0; Whereas X2 < X ≤ X3.

Therefore the first test of the maximum points which needs that the first derivative of a function equals zero or

(dy)/(dx) = f'(x) < 0

The second text of a maximum point which needs that the second derivative of a function is negative or

            (d²y)/(dx²) = f''(x) < 0

Illustration

Find out the critical value for the given functions and determine the critical value that constitutes a maximum

            y = x³ - 12x² + 36x + 8

Solution

            y = x³ - 12x² + 36x + 8

            Then    (dy)/(dx) = 3x² - 24x + 36 +0

i.                    The critical values for the function are acquired by equating the first derivative of the function to zero, which is as:

(dy)/(dx) =  0 or 3x² - 24x + 36 = 0

Thus (x-2) (x-6) = 0

And also x = 2 or 6

The critical values for x are x = 2 or may be 6 and critical values for the function are y = 40 or maybe 8

i.                    To ascertain where these critical values of x will provides rise to a maximum that we apply the second text, which is :

(d²y)/( d²x)       < 0

(dy)/(dx) = 3x² - 24x + 36 and

(d²y)/(d²x) = 6x - 24

a)      When x = 2

Then    (d²y)/( d²x)    = -12 <0

b)      When x = 6

Then    (d²y)/( d²x)    = +2 > 0

Therefore a maximum occurs when x = 2, as this value of x satisfies the second condition. X = 6 does not provides rise to a local maximum