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The critical points of f(x)=x3+x2+x+1f(x) = x^3 + x^2 + x + 1 are

Solution

✅ Correct Option: 4

f′(x)=3x2+2x+1f'(x) = 3x^2 + 2x + 1. For critical points, set f′(x)=0f'(x)=0. Discriminant =4−12=−8<0= 4 - 12 = -8 < 0, so no real roots. Hence the critical points do not exist.

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