Find the intervals in which the following functions are strictly increasing or decreasing.
(a) x2 + 2x - 5
(b) 10 - 6x - 2x2
(c) - 2x3 - 9x2 - 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) = x2 + 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 - 2x2
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) = - 2x3 - 9x2 - 12x + 1
Hence,
f' (x) = - 6x2 - 18x - 12
= - 6 (x2 + 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 - x2
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) x2 + 2x - 5 (b) 10 - 6x - 2x2 (c) - 2x3 - 9x2 - 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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