Question

A function $y=f\left(x\right)$ has a second order derivative $f^{\prime \prime }=6(x-1)$. 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:

• A. $(x-1{)}^2$
• B. $(x-1{)}^3$
• C. $(x+1{)}^3$
• D. $(x+1{)}^2$

Answer 1

The correct option is B $(x-1{)}^3$

$f"\left(x\right)=6(x-1)\Rightarrow f^{\prime }\left(x\right)=3{(x-1)}^2+c$ ...(1)
At the point $(2,1)$ the tangent to graph is $y=3x-5$
Slope of tangent $=3$
$\therefore f^{\prime }\left(2\right)=3{(2-1)}^2+c=3\Rightarrow 3+c=3\Rightarrow c=0$
$\therefore f^{\prime }\left(x\right)=3{(x-1)}^2\Rightarrow f\left(x\right)={(x-1)}^3+k$
Since graph passes through $(2,1)$
$1={(2-1)}^3+k\Rightarrow k=0$
Therefore equation of function is $f\left(x\right)={(x-1)}^3$