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