The total number of extremum point(s) for f(x)=x3+x2+x+1 is
A
0.000
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B
0.0
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C
0
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D
0.00
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Solution
Given, f(x)=x3+x2+x+1 ⇒f′(x)=3x2+2x+1 D=4−12=−8<0
Hence, f′(x)>0∀x∈R
So, There is no critical points for given f(x) ⇒ There is no possible point of local maxima/minima.