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Question

Let f(x) be a polynomial of degree 3 such that f(1)=10,f(1)=6,f(x) has a critical point at x=1 and f(x) has a critical point at x=1. Then f(x) has a local minima at x equals to

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Solution

Let the polynomial be
f(x)=ax3+bx2+cx+d
f(x)=3ax2+2bx+c
f′′(x)=6ax+2b
f′′(1)=06a+2b=0b=3a
f(1)=03a2b+c=0
c=9a
f(1)=10a+bc+d=10
a3a+9a+d=10
d=5a+10
f(1)=6a+b+c+d=6
a3a9a5a+10=6
a=1
f(x)=3x26x9=3(x22x3)
For f(x)=0x22x3=0x=3,1
Minima exists at x=3

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