# Find the intervals in which the following functions are strictly increasing or decreasing.

(a) x^{2} + 2x - 5

(b) 10 - 6x - 2x^{2}

(c) - 2x^{3} - 9x^{2} - 12x + 1

(d) 6 - 9x - 9x2

(e) (x + 1)^{3} (x - 3)^{3}

**Solution:**

Increasing functions are those functions that increase monotonically within a particular domain,

and decreasing functions are those which decrease monotonically within a particular domain.

(a) f (x) = x^{2} + 2x - 5

Hence,

f' (x) = 2x + 2

Therefore,

⇒ f' (x) = 0

⇒ x = - 1

x = - 1 divides the number line into intervals (- ∞, 1) and (- 1, ∞)

In (- ∞, 1),

f' (x) = 2x + 2 < 0

Thus, f is strictly decreasing in (- ∞, 1)

In (- 1, ∞),

f' (x) = 2x + 2 > 0

Thus, f is strictly increasing in (- 1, ∞)

(b) f (x) = 10 - 6x - 2x^{2}

Hence,

f' (x) = - 6 - 4x

Therefore,

⇒ f' (x) = 0

⇒ x = - 3/2

x = - 3/2, divides the number line into two intervals (- ∞, 3/2) and (- 3/2, ∞)

In (- 3/2, ∞),

f' (x) = - 6 - 4x < 0

Hence, f is strictly increasing for x < - 3/2

In (- 3/2, ∞),

f' (x) = - 6 - 4x > 0

Hence, f is strictly increasing for x > - 3/2

(c) f (x) = - 2x^{3} - 9x^{2} - 12x + 1

Hence,

f' (x) = - 6x^{2} - 18x - 12

= - 6 (x^{2} + 3x + 2)

= - 6 (x + 1)(x + 2)

Therefore,

⇒ f' (x) = 0

⇒ x = - 1, 2

x = - 1 and x = - 2 divide the number line into intervals (- ∞, - 2), (- 2, - 1) and (- 1, ∞)

In (- ∞, - 2) and (- 1, ∞),

f' (x) = - 6 (x + 1)(x + 2) < 0

Hence, f is strictly decreasing for x < - 2 and x > - 1

In (- 2, - 1),

f' (x) = - 6 ( x + 1)(x + 2) > 0

Hence, f is strictly increasing in - 2 < x < - 1

(d) f (x) = 6 - 9x - x^{2}

Hence,

f' (x) = - 9 - 2x

Therefore,

⇒ f' (x) = 0

⇒ x = - 9/2

In (- 9/2, ∞),

f' (x) < 0

Hence, f is strictly decreasing for x > - 9/2

In (- ∞, - 9/2), f' (x) > 0

Hence, f is strictly decreasing in x > - 9/2

(e) (x + 1)^{3} (x - 3)^{3}

Hence,

f' (x) = 3(x + 1)^{2} (x - 3)^{3} + 3(x - 3)^{2} (x + 1)^{3}

= 3(x + 1)^{2} (x - 3)^{2} [x - 3 + x + 1]

= 3(x + 1)^{2} (x - 3)^{2} (2x - 2)

= 6 (x + 1)^{2} (x - 3)^{2} (x - 1)

Therefore,

⇒ f' (x) = 0

⇒ x = - 1, 3, 1

x = - 1, 3, 1 divides the number line into four intervals (- ∞, - 1), (- 1, 1), (1, 3) and (3, ∞)

In (- ∞, - 1) and (- 1, 1) , f' (x) = 6 (x + 1)^{2} (x - 3)^{2} (x - 1) < 0

Hence, f is strictly decreasing in (- ∞, - 1) and (- 1, 1)

In (1, 3) and (3, ∞),

f' (x) = 6 (x + 1)^{2} (x - 3)^{2} (x - 1) > 0

Hence, f is strictly increasing in (1, 3) and (3, ∞)

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.2 Question 6

## Find the intervals in which the following functions are strictly increasing or decreasing. (a) x^{2} + 2x - 5 (b) 10 - 6x - 2x^{2 }(c) - 2x^{3} - 9x^{2} - 12x + 1 (d) 6 - 9x - 9x2 (e) (x + 1)^{3} (x - 3)^{3}

**Summary:**

The intervals in which the following functions are strictly increasing or decreasing is (a) f is strictly increasing in (- 1, ∞) and f is strictly decreasing in (- ∞, 1) (b) f is strictly increasing for x < - 3/2 and f is strictly increasing for x > - 3/2

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