Find the intervals in which the following functions are strictly increasing or decreasing:
\((a) x^2 + 2x − 5 \)
\((b) 10 − 6x − 2x^2 \)
\((c) −2x^3 − 9x^2 − 12x + 1 \)
\((d) 6 − 9x − x^2 \)
\((e) (x + 1)^3 (x − 3)^3\)
(a) We have,
f(x)=x2+2x-5
∴ f'(x)=2x+2
Now,
f'(x)=0=x=-1
Point x = −1 divides the real line into two disjoint intervals i.e.,(-∞,-1) and (-1.∞)
In interval (-∞,-1),f'(x)=2x+2<0.
∴f is strictly decreasing in interval (-∞,-1).
Thus, f is strictly decreasing for x < −1.
(b) In interval (-1,∞), f'(x) = 2x+2>0.
∴ f is strictly increasing in interval (-1,∞).
Thus, f is strictly increasing for x > −1. (b) We have,
f(x) = 10 − 6x − 2x2
=f'(x)=-6-4x
Now,
f'(x) = 0 = x = -\(\frac 32\)
The point x=-\(\frac 32\) divides the real line into two disjoint intervals
i.e.,(-∞,-\(\frac 32\)) and (-\(\frac 32\),∞).
In interval (-∞,-\(\frac 32\)) i.e., when x<-\(\frac 32\), f'(x)=-6-4x<0.
∴ f is strictly increasing for x<-\(\frac 32\).
In interval i.e., (-∞,-\(\frac 32\)) when x<-\(\frac 32\), f'(x)=-6-4x<0.
∴ f is strictly increasing for x<-\(\frac 32\).
In interval i.e., (-∞,-\(\frac 32\)) when x>-\(\frac 32\), f'(x)=-6-4x<0.
∴ f is strictly increasing for x>-\(\frac 32\).
(c) We have, f(x) = −2x3 − 9x2 − 12x + 1
f'(x)=-6x2-18x-12=-6(x2+3x+2)=-6(x-1)(x+2)
Now,
f'(x)=0=x=-1 and x=-2
Points x = −1 and x = −2 divide the real line into three disjoint intervals
i.e.,(-∞,-2),(-2,-1), and (-1,∞).
In intervals (-∞,-2) and (-1,∞) i.e., when x<−2 and x>−1,
f'(x) = -6(x-1)(x+2)<0.
∴ f is strictly decreasing for x<−2 and x>−1.
Now, in interval (−2,−1) i.e., when −2<x<−1, .f'(x)=-6(x+1)(x+2)>0
∴ f is strictly increasing for -2<x<-1.
(d) We have,
f(x)=6-9x-x2
∴f'(x)=-9-2x
Now, f'
(x)=0 gives x=-\(\frac 92\)
The point x=-\(\frac 92\) divides the real line into two disjoint intervals i.e.,
(-∞,-\(\frac 92\)) and (\(\frac 92\),∞).
In interval (-∞,-\(\frac 92\)) i.e., for x<-\(\frac 92\), f'(x) = -9-2x>0.
∴ f is strictly increasing for x<-\(\frac 92\).
In interval i.e., (-\(\frac 92\),∞) for x>-\(\frac 92\), f'(x) = -9-2x<0.
∴ f is strictly decreasing for x>-\(\frac 92\).
(e) We have, f(x) = (x + 1)3 (x − 3)3
f'(x) = 3(x+1)2(x-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)
Now,
f'(x)=0=x=-1,3,1
The points x = −1, x = 1, and x = 3 divide the real line into four disjoint intervals
i.e.,(-∞,-1),(-1,1)(1,3) and(3,∞).
In intervals (-∞,-1) and (-1,1), f'(x)=6(x+1)2(x-3)2(x-1)<0.
∴ f is strictly decreasing in intervals (-∞,-1) and (−1, 1).
In intervals (1, 3) and (3,∞), f'(x)=6(x+1)2(x-3)2(x-1)>0.
∴ f is strictly increasing in intervals (1, 3) and (3,∞).
If \( x = a(0 - \sin \theta) \), \( y = a(1 + \cos \theta) \), find \[ \frac{dy}{dx}. \]
Find the least value of ‘a’ for which the function \( f(x) = x^2 + ax + 1 \) is increasing on the interval \( [1, 2] \).
Arrange the following states in sequence (highest to lowest) according to their reserves of iron ore and choose the correct option.
I. Jharkhand
II. Karnataka
III. Chhattisgarh
IV. Odisha
Increasing Function:
On an interval I, a function f(x) is said to be increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≤ f(y)
Decreasing Function:
On an interval I, a function f(x) is said to be decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) ≥ f(y)
Strictly Increasing Function:
On an interval I, a function f(x) is said to be strictly increasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) < f(y)
Strictly Decreasing Function:
On an interval I, a function f(x) is said to be strictly decreasing, if for any two numbers x and y in I such that x < y,
⇒ f(x) > f(y)