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Question

Let f(x) be a cubic polynomial with f(1)=10, f(1)=6, and has a local minima at x=1, and f(x) has a local minima at x=1. Then f(3) is equal to

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Solution

f(1)=10, f(1)=6
f(1)=0 and f′′(1)=0 as given
f(x) has minima at x=1
and f(x) has minima at x=1
So, f′′(x)=a(x+1)
Integrating both sides
f(x)=a2(x+1)2+c
f(1)=0=2a+c
c=2a
f(x)=a2(x+1)22a
Intergrating both side
f(x)=a6(x+1)32xa+c
f(1)=10=8a62a+c
f(1)=6=2a+c
2a+c=6
4a6c=60
a=6,c=6
f(x)=(x+1)312x6
f(3)=(4)3366
f(3)=22

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