Question

Find the intervals in which the following functions are strictly increasing or decreasing: 
\((a) x^2 + 2x − 5 \)
\((b) 10 − 6x − 2x^2 \)
\((c) −2x^3 − 9x^2 − 12x + 1 \)
\((d) 6 − 9x − x^2 \)
\((e) (x + 1)^3 (x − 3)^3\)

Answer 1

(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.

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(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>-\(\frac 32\).

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(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.

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(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>-\(\frac 92\).

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(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,∞).