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