Problem

Problem

Statement I The function f(x)=(x3+3x−4)(x2+4x−5)f(x) = (x^3 + 3x - 4)(x^2 + 4x - 5) has local extremum at x=1x = 1. Statement II f(x)f(x) is continuous and differentiable and f′(1)=0f'(1) = 0.
Intuitive explanation

To determine if f(x)f(x) has a local extremum at x=1x = 1, you need to check whether the derivative equals zero there and whether the derivative actually changes sign around that point. Statement II tells you that f′(1)=0f'(1) = 0, which is necessary but not sufficient—a critical point is only a local extremum if the function changes from increasing to decreasing (or vice versa). The key is to examine the sign of f′(x)f'(x) on either side of x=1x = 1 using the product rule, or to check the second derivative test, to confirm whether this critical point is truly an extremum.

Intuitive explanation

To determine if f(x)f(x) has a local extremum at x=1x = 1, you need to check whether the derivative equals zero there and whether the derivative actually changes sign around that point. Statement II tells you that f′(1)=0f'(1) = 0, which is necessary but not sufficient—a critical point is only a local extremum if the function changes from increasing to decreasing (or vice versa). The key is to examine the sign of f′(x)f'(x) on either side of x=1x = 1 using the product rule, or to check the second derivative test, to confirm whether this critical point is truly an extremum.

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