# What is the cubic function that has a minimum value at the point (2;4) and the maximum value at (5,1)?

giorgiana1976 | Student

Let the cubic function be f(x) = ax^3 + bx^2 + cx +d

We know that the x coordinates of the local extrema are the zeroes of the 1st derivative of the function.

We'll differentiate with respect to x to determine the 1st derivative:

f'(x) = 3ax^2 + 2bx + c

f'(2) = 0 => f'(2) = 12a + 4b + c = 0 (1)

f'(5) = 0 => f'(5) = 75a + 10b + c = 0 (2)

We also know that the coordinates (2,4) and (5,1) verify the expression of the function;

f(2)=4 <=> 8a + 4b + 2c + d = 4 (3)

f(5) = 1 <=> 125 a + 25b + 5c + d = 1 (4)

We'll subtract (1) from (2):

63a + 6b = 0 => 21a + 2b = 0 (5)

We'll subtract (3) from (4):

117a + 21b + 3c = -3 (6)

We'll multiply (1) by -3:

-36a - 12b - 3c = 0 (7)

-36a - 12b - 3c + 117a + 21b + 3c = -3

81a + 9b = -3

27a + 3b = -1 (8)

21a + 2b = 0 => b = -21a/2

27a  - 63a/2 = -1

54a - 63a = -2

-9a = -2 => a = 2/9 => b = -21*2/2*9 => b = -7/3

12a + 4b + c = 0

12*2/9 - 28/3 + c = 0

24 - 84 + 9c = 0

-9c = -60 => c = 20/3

8a + 4b + 2c + d = 4

16/9 - 28/3 + 40/3 + d = 4

16 - 84 - 120 - 36 = -9d

-224 = -9d => d = 224/9

The cubic function is f(x) = (1/3)(2x^3/3 -7x^2 + 20x + 224/3)