The second derivative of the function y = f(x | vedclass

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(B) Given $f''(x) = 6(x - 1)$.Integrating with respect to $x$,we get $f'(x) = \int 6(x - 1) \, dx = 3(x - 1)^2 + C_1$.Since the tangent at $(2, 1)$ is

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$(x - 1)^3$

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(B) Given $f''(x) = 6(x - 1)$.Integrating with respect to $x$,we get $f'(x) = \int 6(x - 1) \, dx = 3(x - 1)^2 + C_1$.Since the tangent at $(2, 1)$ is $y = 3x - 5$,the slope of the tangent at $x = 2$ is $f'(2) = 3$.Substituting $x = 2$ into the expression for $f'(x)$,we get $f'(2) = 3(2 - 1)^2 + C_1 = 3 + C_1$.Equating this to $3$,we have $3 + C_1 = 3$,which implies $C_1 = 0$.Thus,$f'(x) = 3(x - 1)^2$.Integrating again,$f(x) = \int 3(x - 1)^2 \, dx = (x - 1)^3 + C_2$.Since the graph passes through $(2, 1)$,we have $f(2) = 1$.Substituting $x = 2$ into the expression for $f(x)$,we get $f(2) = (2 - 1)^3 + C_2 = 1 + C_2$.Equating this to $1$,we have $1 + C_2 = 1$,which implies $C_2 = 0$.Therefore,the function is $f(x) = (x - 1)^3$.

The second derivative of the function $y = f(x)$ is $f''(x) = 6(x - 1)$. If the graph of the function passes through the point $(2, 1)$ and the equation of the tangent at that point is $y = 3x - 5$,find the equation of the function.

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