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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

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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 $3 = 3(2 - 1)^2 + c_1$,which implies $3 = 3 + c_1$,so $c_1 = 0$.Thus,$f'(x) = 3(x - 1)^2$.Integrating again with respect to $x$,we get $f(x) = \int 3(x - 1)^2 dx = (x - 1)^3 + c_2$.Since the graph passes through $(2, 1)$,we have $f(2) = 1$.Substituting $x = 2$ into the expression for $f(x)$,we get $1 = (2 - 1)^3 + c_2$,which implies $1 = 1 + c_2$,so $c_2 = 0$.Therefore,the function is $f(x) = (x - 1)^3$.

$A$ function $y = f(x)$ has a second-order derivative $f''(x) = 6(x - 1)$. If its graph passes through the point $(2, 1)$ and at that point the tangent to the graph is $y = 3x - 5$,then the function is

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