Question

Antral gyat kijiye (X+1)^3(x-3)^3=?

Answer 1

Step-by-step explanation:

Text Solution

Solution :

f(x)=(x+1)3(x−3)3

⇒f'(x)=(x+1)3.ddx(x−3)3+(x−3)3.ddx(x+1)3

=(x+1)3.3(x−3)2+(x−3)3.3(x+1)2

=3(x+1)2(x−3)2[(x+1)+(x−3)]

=6(x+1)2(x−3)2(x−1)...(i)

(a) f(x) is increasing

⇔f'(x)≥0

⇔6(x+1)2(x−3)2(x−1)≥0 [from (i)]

⇔(x−1)≥0

⇔x≥1

⇔x∈[1,∞]

∴f(x) is increasing on [1,∞]

(b) f(x) is decreasing

⇔f'(x)≤0

⇔6(x+1)2(x−3)2(x−1)≤0 [from (i)]

⇔(x−1)≤0⇔x≤1

⇔x∈[−∞,1]

∴f(x) is decreasing on [−∞,1]

Hence, f(x) is increasing on [1,∞] and decreasing on [−∞,1]