In this article, we shall study the concept of maximum and minimum values of a function using the concept of differentiation.
Example – 01:
Find the maximum and minimum values of x3 – 12x – 5
Solution:
Let ƒ(x) = x3 – 12x – 5 ………….. (1)
Differentiating equation (1) w.r.t. x
ƒ’(x) = 3x2 – 12 ………….. (2)
Differentiating equation (2) w.r.t. x
ƒ’’(x) = 6x ………….. (3)
For maximum or minimum value ƒ’(x) = 0
3x2 – 12 = 0
∴ 3(x2 – 4) = 0
∴ 3(x + 2)(x – 2) = 0
∴ x + 2 = 0 or/and x – 2 = 0
∴ x = – 2 or/and x = 2
Let us consider x = 2
ƒ’’(2) = 6 x 2 = 12 > 0
Hence the function is increasing at x = 2 and has minimum value at x = 2
Substituting x = 2 in equation (1)
Minimum value = ƒ(2) = (2)3 – 12(2) – 5 = 8 – 24 – 5 = – 21
Thus point of minimum is (2, -21)
Let us consider x = – 2
ƒ’’(-2) = 6 x (- 2) = – 12 < 0
Hence the function is decreasing at x = – 2 and has maximum value at x = -2
Substituting x = – 2 in equation (1)
Maximum value = ƒ(-2) = (-2)3 – 12(-2) – 5 = – 8 + 24 – 5 = 11
Thus point of maximum is (- 2, 11)
Ans: The maximum value is 11 at x = -2 and minimum value is – 21 at x = 2
Example – 02:
Find the maximum and minimum values of x3 – 9x2 + 24 x
Solution:
Let ƒ(x) = x3 – 9x2 + 24 x ………….. (1)
Differentiating equation (1) w.r.t. x
ƒ’(x) = 3x2 – 18x + 24 ………….. (2)
Differentiating equation (2) w.r.t. x
ƒ’’(x) = 6x – 18 ………….. (3)
For maximum or minimum value ƒ’(x) = 0
3x2 – 18x + 24 = 0
∴ x2 – 6x + 8 = 0
∴ (x – 4)(x – 2) = 0
∴ x – 4 = 0 or/and x – 2 = 0
∴ x = 4 or/and x = 2
Let us consider x = 4
ƒ’’(4) = 6 x 4 – 18 = 24 – 18 = 6 > 0
Hence the function is increasing at x = 4 and has minimum value at x = 4
Substituting x = 4 in equation (1)
Minimum value = ƒ(4) = (4)3 – 9(4)2 + 24(4) = 64 – 144 + 96 = 16
Thus point of minimum is (2, -21)
Let us consider x = – 2
ƒ’’(-2) = 6 x (- 2) – 18 = -12 – 18 = – 30 < 0
Hence the function is decreasing at x = 2 and has maximum value at x = 2
Substituting x = 2 in equation (1)
Maximum value = ƒ(2) = (2)3 – 9(2)2 + 24(2) = 8 – 36 + 48 = 20
Thus point of maximum is (2, 20)
Ans: The maximum value is 20 at x = 2 and minimum value is 16 at x = 4
Example – 03:
Find the maximum and minimum values of 2x3 – 3x2 – 36x + 10
Solution:
Let ƒ(x) = 2x3 – 3x2 – 36x + 10 ………….. (1)
Differentiating equation (1) w.r.t. x
ƒ’(x) = 6x2 – 6x – 36 ………….. (2)
Differentiating equation (2) w.r.t. x
ƒ’’(x) = 12x – 6 ………….. (3)
For maximum or minimum value ƒ’(x) = 0
6x2 – 6x – 36 = 0
∴ x2 – x – 6 = 0
∴ (x – 3)(x + 2) = 0
∴ x – 3 = 0 or/and x + 2 = 0
∴ x = 3 or/and x = – 2
Let us consider x = 3
ƒ’’(3) = 12(3) – 6 = 36 – 6 = 30 > 0
Hence the function is increasing at x = 3 and has minimum value at x = 3
Substituting x = 3 in equation (1)
Minimum value = ƒ(3) = 2(3)3 – 3(3)2 – 36(3) + 10 = 54 – 27 – 108 = -71
Thus point of minimum is (3, -71)
Let us consider x = – 2
ƒ’’(2) = 12(-2) – 6 = – 24 – 6 = – 30 < 0
Hence the function is decreasing at x = – 2 and has maximum value at x = – 2
Substituting x = – 2 in equation (1)
Maximum value = ƒ(-2) = 2(-2)3 – 3(-2)2 – 36(-2) + 10 = – 16 – 12 + 72 + 10 = 54
Thus point of maximum is (-2, 54)
Ans: The maximum value is 54 at x = – 2 and minimum value is – 71 at x = 3
Example – 04
Find the maximum and minimum values of x3 + 3x2 – 2
Solution:
Let ƒ(x) = x3 + 3x2 – 2 ………….. (1)
Differentiating equation (1) w.r.t. x
ƒ’(x) = 3x2 + 6x ………….. (2)
Differentiating equation (2) w.r.t. x
ƒ’’(x) = 6x + 6 ………….. (3)
For maximum or minimum value ƒ’(x) = 0
3x2 + 6x = 0
∴ 3x(x + 2) = 0
∴ x = 0 or/and x + 2 = 0
∴ x = 0 or/and x = – 2
Let us consider x = 0
ƒ’’(-2) = 6(0) + 6 = 6 > 0
Hence the function is increasing at x = 0 and has minimum value at x = 0
Substituting x = 0 in equation (1)
Minimum value = ƒ(0) = (0)3 + 3(0)2 – 2 = 0 + 0 – 2 = -2
Thus point of minimum is (0, -2)
Let us consider x = – 2
ƒ’’(-2) = 6(-2) + 6 = -12 + 6 = – 6 < 0
Hence the function is decreasing at x = – 2 and has maximum value at x = – 2
Substituting x = – 2 in equation (1)
Maximum value = ƒ(-2) = (-2)3 + 3(-2)2 – 2 = – 8 + 12 – 2 = 2
Thus point of maximum is (-2, 2)
Ans: The maximum value is 2 at x = – 2 and minimum value is – 2 at x = 0
Example – 05:
Find the maximum and minimum values of 3x3 – 9x2 – 27x + 15
Solution:
Let ƒ(x) = 3x3 – 9x2 – 27x + 15 ………….. (1)
Differentiating equation (1) w.r.t. x
ƒ’(x) = 9x2 – 18x – 27 ………….. (2)
Differentiating equation (2) w.r.t. x
ƒ’’(x) = 18x – 18 ………….. (3)
For maximum or minimum value ƒ’(x) = 0
9x2 – 18x – 27 = 0
∴ x2 – 2x – 3 = 0
∴ (x – 3)(x + 1) = 0
∴ x – 3 = 0 or/and x + 1 = 0
∴ x = 3 or/and x = – 1
Let us consider x = 3
ƒ’’(3) = 18(3) – 18 = 54 – 18 = 36 > 0
Hence the function is increasing at x = 3 and has minimum value at x = 3
Substituting x = 3 in equation (1)
Minimum value = ƒ(3) = 3(3)3 – 9(3)2 – 27(3) + 15 = 81 – 81 -81 +15 = – 66
Thus point of minimum is (3, – 66)
Let us consider x = – 1
ƒ’’(2) = 18(-1) – 18 = – 18 – 18 = – 36 < 0
Hence the function is decreasing at x = – 1 and has maximum value at x = – 1
Substituting x = – 1 in equation (1)
Maximum value = ƒ(-1) = 3(-1)3 – 9(-1)2 – 27(-1) + 15 = – 3 – 9 + 27 + 15 = 30
Thus point of maximum is (-1, 30)
Ans: The maximum value is 30 at x = – 1 and minimum value is – 66 at x = 3
Example – 06:
Find the maximum and minimum values of 2x3 – 21x2 + 36x – 20
Solution:
Let ƒ(x) = 2x3 – 21x2 + 36x – 20 ………….. (1)
Differentiating equation (1) w.r.t. x
ƒ’(x) = 6x2 – 42x + 36 ………….. (2)
Differentiating equation (2) w.r.t. x
ƒ’’(x) = 12x – 42 ………….. (3)
For maximum or minimum value ƒ’(x) = 0
6x2 – 42x + 36 = 0
∴ x2 – 7x + 6 = 0
∴ (x – 6)(x – 1) = 0
∴ x – 6 = 0 or/and x – 1 = 0
∴ x = 6 or/and x = 1
Let us consider x = 6
ƒ’’(6) = 12(6) – 42 =72 – 42 = 30 > 0
Hence the function is increasing at x = 6 and has minimum value at x = 6
Substituting x = 6 in equation (1)
Minimum value = ƒ(3) = 2(6)3 – 21(6)2 + 36(6) – 20 = 432 – 756 = 216 – 20 = – 128
Thus point of minimum is (6, – 128)
Let us consider x = 1
ƒ’’(2) = 12(1) – 42 = 12 – 42 = – 30 < 0
Hence the function is decreasing at x = 1 and has maximum value at x = 1
Substituting x = 1 in equation (1)
Maximum value = ƒ(1) = 2(1)3 – 21(1)2 + 36(1) – 20 = 2 – 21 +36 – 20 = – 3
Thus point of maximum is (1, -3)
Ans: The maximum value is -3 at x = 1 and minimum value is – 128 at x = 6
Example – 07:
Find the maximum and minimum values of x2 + 16/x2
Solution:
Let ƒ(x) = x2 + 16/x2
ƒ(x) = x2 + 16 x -2 ………….. (1)
Differentiating equation (1) w.r.t. x
ƒ’(x) = 2x – 32x -3 ………….. (2)
Differentiating equation (2) w.r.t. x
ƒ’’(x) = 2 + 96 x -4 ………….. (3)
For maximum or minimum value ƒ’(x) = 0
2x – 32x -3 = 0
∴ 2x = 32x -3
∴ 2x = 32/x3
∴ x4 = 16
∴ x = ± 2
Let us consider x = 2
ƒ’’(2) = 2 + 96 x (2) -4 = 2 + 96/16 = 2 + 6 = 8 > 0
Hence the function is increasing at x = 2 and has minimum value at x = 2
Substituting x = 2 in equation (1)
Minimum value = ƒ(2) = 22 + 16/22 = 4 + 4 = 8
Thus point of minimum is (2, 8)
Let us consider x = – 2
ƒ’’(2) = – 2 + 96 x (- 2) -4 = – 2 + 96/16 = – 2 + 6 = 4 > 0
Hence the function is increasing at x = – 2 and has minimum value at x = – 2
Substituting x = – 2 in equation (1)
Minimumm value = ƒ(-2) =(-2)2 + 16/(-2)2 = 4 + 4 = 8
Thus point of minimum is (- 2, 8)
Ans: The minimum value is 8 at x = ±2
Example – 08:
Find the maximum and minimum values of x.logx
Solution:
Let ƒ(x) = x.logx ………….. (1)
Differentiating equation (1) w.r.t. x
ƒ’(x) = x(1/x) + logx. (1)
ƒ’(x) = 1 + logx ………….. (2)
Differentiating equation (2) w.r.t. x
ƒ’’(x) = 1/x ………….. (3)
For maximum or minimum value ƒ’(x) = 0
1 + logx = 0
∴ log x = -1
∴ x = e-1 = 1/e
ƒ’’(1/e) = 1/(1/e) = e > 0
Hence the function is increasing at x = 1/e and has minimum value at x = 1/e
Substituting x = 1/e in equation (1)
Minimum value = ƒ(1/e) = (1/e). log(1/e) = – (1/e). log(e) = – (1/e)
Thus point of minimum is (1/e, – 1/e)
Ans: The minimum value is – 1/e at x = 1/e