Question

Find the maximum and minimum of the following functions : f(x) = 2x³ – 21x² + 36x – 20

Answer 1

f(x) = 2x³ – 21x² + 36x – 20

∴ f'(x) = `d/dx(2x^3 - 21x^2 + 36x - 20)`

= 2 x 3x² – 21 x 2x + 36 x 1 – 0
= 6x² – 42x + 36
and
f'(x) = `d/dx(6x^2 - 42x + 36)`

= 6 x 2x – 42 x 1 + 0
= 12 x – 42
f'(x) = 0 gives 6x² – 42x + 36 = 0
∴ x2 – 7x + 6 = 0
∴ (x – 1)(x – 6) = 0
∴ 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
= f(1)
= 2(1)³ – 21(1)² + 36(1) – 20
= 2 – 21 + 36 – 20
= – 3

(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
= 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.

Method 2 (Second 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(16) – 20
= 432 – 756 + 216 – 20
= – 128
Hence, the function f has maximum value – 3 at=1 and minimum value – 128 at x = 6.

Note : Out of the two methods, given above, we will use the second derivative test for the remaining problems.