Question

Find the maximum and minimum of the following functions:

f(x) = 2x³ – 21x² + 36x – 20

Answer 1

f(x) = 2x³ – 21x² + 36x – 20

the roots of f'(x) = 0 are x₁ = 1 and x₂ = 6.

Method 1 (Second Derivative Test):

(a) f”(1) = 12(1) – 42 = -30 < 0

∴ by the second derivative test, f has maximum at x = 1 and maximum value of f at x = 1

(b) f”(6) = 12(6) – 42 = 30 > 0

∴ by the second derivative test, f has minimum at x = 6 and minimum value of f at x = 6

Hence, the function f has maximum value -3 at x = 1 and minimum value -128 at x = 6.

Method 2 (First Derivative Test):

(a) f'(x) = 6(x – 1)(x – 6)

Consider x = 1 

Let h be a small positive number. Then

f'(1 – h) = 6(1 – h – 1)(1 – h – 6) 

= 6(-h)(-5 – h) 

= 6h(5 + h)> 0 and f'(1 + h) = 6(1 + h – 1)(1 + h – 6) 

= 6h(h – 5) < 0, as h is small positive number. 

∴ by the first derivative test, f has maximum at x = 1 and maximum value of f at x = 1 

f(1) = 2(1)³ – 21(1)² + 36(1) – 20 

= 2 – 21 + 36 – 20 

= -3

(b) f'(x) = 6(x – 1)(x – 6) 

Consider x = 6 

Let h be a small positive number. 

Then f'(6 – h) = 6(6 – h – 1)(6 – h – 6) 

= 6(5 – h)(-h)

= -6h(5 – h) < 0, as h is small positive number and f'(6 + h) = 6(6 + h – 1)(6 + h – 6) = 6(5 + h)(h) > 0

∴ by the first derivative test, f has minimum at x = 6

and minimum value of f at x = 6 

f(6) = 2(6)³ – 21(6)² + 36(6) – 20 

= 432 – 756 + 216 – 20 

= -128

Hence, the function f has maximum value -3 at x = 1 and minimum value -128 at x = 6.