For the function f(x)=x^3-6x^2+12x-3,the poin | vedclass

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(B) Given the function $f(x)=x^3-6x^2+12x-3$. \ First,find the first derivative: $f'(x)=3x^2-12x+12$. \ Set $f'(x)=0$ to find critical points: $3(x^

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$A$ point of inflexion

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(B) Given the function $f(x)=x^3-6x^2+12x-3$. \ First,find the first derivative: $f'(x)=3x^2-12x+12$. \ Set $f'(x)=0$ to find critical points: $3(x^2-4x+4)=0 \Rightarrow 3(x-2)^2=0 \Rightarrow x=2$. \ Next,find the second derivative: $f''(x)=6x-12$. \ Evaluate the second derivative at $x=2$: $f''(2)=6(2)-12=0$. \ Since $f''(2)=0$,we check the third derivative: $f'''(x)=6$. \ Since $f'''(2)=6 
eq 0$,the concavity changes at $x=2$. \ Therefore,$x=2$ is a point of inflexion.

For the function $f(x)=x^3-6x^2+12x-3$,the point $x=2$ is

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