Question

Minimum value of f(x) = 6x^3 - 45x^2 +108x + 2/ 2x^3 - 15x^2 + 36x + 1 will occur when x is equal to

Answer 1

Answer:

x = 3

Step-by-step explanation:

Minimum value of f(x) = 6x^3 - 45x^2 +108x + 2/ 2x^3 - 15x^2 + 36x + 1 will occur when x is equal to

f(x) = (6x^3 - 45x^2 +108x + 2)/ (2x^3 - 15x^2 + 36x + 1)

=> f(x) = (6x^3 - 45x^2 +108x + 3 - 1)/ (2x^3 - 15x^2 + 36x + 1)

=> f(x) = (3(2x^3 - 15x^2 +36x + 1) - 1)/ (2x^3 - 15x^2 + 36x + 1)

=> f(x) = 3 - 1/(2x^3 - 15x^2 + 36x + 1)

f(x) would be minimum if

2x^3 - 15x^2 + 36x + 1 is + ve & least

g(x) = 2x^3 - 15x^2 + 36x + 1

g'(x) = 6x² - 30x + 36

g''(x) = 12x - 30

6x² - 30x + 36 = 0

=> x² - 5x + 6 = 0

=> x = 2 & 3

g''(x) is -ve if x = 2

g''(x) is +ve if x = 3

x = 3