Let f(x) = x^3 - 6x^2 + 12x - 3. Then at x = | vedclass

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(D) To determine the nature of the function at $x = 2$,we find the first and second derivatives of $f(x) = x^3 - 6x^2 + 12x - 3$.Step $1$: Find the fi

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neither a maximum nor a minimum

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(D) To determine the nature of the function at $x = 2$,we find the first and second derivatives of $f(x) = x^3 - 6x^2 + 12x - 3$.Step $1$: Find the first derivative $f'(x)$.$f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 12x - 3) = 3x^2 - 12x + 12$.Step $2$: Find the critical points by setting $f'(x) = 0$.$3x^2 - 12x + 12 = 0 \implies 3(x^2 - 4x + 4) = 0 \implies 3(x - 2)^2 = 0$.This gives $x = 2$ as the critical point.Step $3$: Find the second derivative $f''(x)$.$f''(x) = \frac{d}{dx}(3x^2 - 12x + 12) = 6x - 12$.Step $4$: Evaluate $f''(x)$ at $x = 2$.$f''(2) = 6(2) - 12 = 12 - 12 = 0$.Since $f''(2) = 0$,the second derivative test is inconclusive. We examine the sign of $f'(x)$ around $x = 2$.$f'(x) = 3(x - 2)^2$. Since $(x - 2)^2$ is always non-negative for all real $x$,$f'(x) \geq 0$ for all $x$.Because the derivative does not change sign as $x$ passes through $2$,$x = 2$ is a point of inflection,not a local maximum or minimum.

Let $f(x) = x^3 - 6x^2 + 12x - 3$. Then at $x = 2$,$f(x)$ has:

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