A function y=f(x) has a second order derivative \[f''\left( x \right) = 6\left( {x - 1} \right).\] 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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Official solution
\[f''\left( x \right) = 6\left( {x - 1} \right).\] Inegrating,
we get \[f'\left( x \right) = 3{x^2} - 6x + c\]
Slope at \[\left( {2,1} \right) = f'\left( 2 \right) = c = 3\]
\[\left[ {\,\,} \right.\] As slope of tangent at (2,1) is 3 \[\left. {\,\,} \right]\]
\[\therefore\] \[f'\left( x \right) = 3{x^2} - 6x + 3 = 3{\left( {x - 1} \right)^2}\]
Inegrating again, we get
\[f\left( x \right) = {\left( {x - 1} \right)^3} + D\]
The curve passes through (2,1)
\[ \Rightarrow 1 = {\left( {2 - 1} \right)^3} + D \Rightarrow D = 0\]
\[\therefore\] \[f\left( x \right) = {\left( {x - 1} \right)^3}\]