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#### Example – 01:

- Find the maximum and minimum values of x
^{3}– 12x – 5 **Solution:**

Let ƒ(x) = x^{3} – 12x – 5 ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 3x^{2} – 12 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 6x ………….. (3)

For maximum or minimum value ƒ’(x) = 0

3x^{2} – 12 = 0

∴ 3(x^{2} – 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 x
^{3}– 9x^{2}+ 24 x **Solution:**

Let ƒ(x) = x^{3} – 9x^{2} + 24 x ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 3x^{2} – 18x + 24 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 6x – 18 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

3x^{2} – 18x + 24 = 0

∴ x^{2} – 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 2x
^{3}– 3x^{2}– 36x + 10 **Solution:**

Let ƒ(x) = 2x^{3} – 3x^{2} – 36x + 10 ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 6x^{2} – 6x – 36 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 12x – 6 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

6x^{2} – 6x – 36 = 0

∴ x^{2} – 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 x
^{3}+ 3x^{2}– 2 **Solution:**

Let ƒ(x) = x^{3} + 3x^{2} – 2 ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 3x^{2} + 6x ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 6x + 6 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

3x^{2} + 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 3x
^{3}– 9x^{2}– 27x + 15 **Solution:**

Let ƒ(x) = 3x^{3} – 9x^{2} – 27x + 15 ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 9x^{2} – 18x – 27 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 18x – 18 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

9x^{2} – 18x – 27 = 0

∴ x^{2} – 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 2x
^{3}– 21x^{2}+ 36x – 20 **Solution:**

Let ƒ(x) = 2x^{3} – 21x^{2} + 36x – 20 ………….. (1)

Differentiating equation (1) w.r.t. x

ƒ’(x) = 6x^{2} – 42x + 36 ………….. (2)

Differentiating equation (2) w.r.t. x

ƒ’’(x) = 12x – 42 ………….. (3)

For maximum or minimum value ƒ’(x) = 0

6x^{2} – 42x + 36 = 0

∴ x^{2} – 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 x
^{2}+ 16/x^{2} **Solution:**

Let ƒ(x) = x^{2} + 16/x^{2 }

ƒ(x) = x^{2} + 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/x^{3}

∴ x^{4} = 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) = 2^{2} + 16/2^{2} = 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

Science > Mathematics > Applications of Derivatives > You are Here |

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