In this page solution of question4 we are going to see detailed solution of first question in the topic maximum and minimum.
Question 4:
Find the maximum and minimum value of the function x³ - 3 x² - 9 x + 12
Solution:
Let y = f (x) = x³ - 3 x² - 9 x + 12
f ' (x) = 3x² - 3 (2x) - 9 (1) + 0
f ' (x) = 3x² - 6x - 9
set f ' (x) = 0
3x² - 6x - 9 = 0
÷ by 3 => x² - 2 x - 3 = 0
(x + 1) (x - 3) = 0
x + 1 = 0 x - 3 = 0
x = -1 x = 3
f ' (x) = 3x² - 6x - 9
f '' (x) = 3 (2 x) - 6 (1) - 0
f '' (x) = 6 x - 6
Put x = -1
f '' (-1) = 6(-1) - 6
= -6 - 6
f '' (-1) = -12 < 0 Maximum
To find the maximum value let us apply x = -1 in the original function
f (x) = x³ - 3 x² - 9 x + 12
f (-1) = (-1)³ - 3 (-1)² - 9 (-1) + 12
= -1 - 3(1) + 9 + 12
= -1 - 3 + 9 + 12
= -4 + 21
= 17
Put x = 3
f '' (3) = 6(3) - 6
= 18 - 6
f '' (3) = 12 > 0 Minimum
To find the minimum value let us apply x = 3 in the original function
f (x) = x³ - 3 x² - 9 x + 12
f (3) = (3)³ - 3 (3)² - 9 (3) + 12
= 27 - 3(9) - 27 + 12
= 27 - 27 - 27 + 12
= -27 + 12
= -15
Therefore themaximum value = 17 and
The minimum value = -15