Question

A function $y=f\left(x\right)$ has a second order derivative $f"\left(x\right)=6(x-1).$ 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 $(x-1{)}^3$

$f"\left(x\right)=6(x-1).$ Integrating,we get
$f^{\prime }\left(x\right)=3x^2-6x+c$
Slope at $(2,1)=f^{\prime }\left(2\right)=c=3$
[$\because$ slope of tangent at (2,1) is 3]
$\therefore f^{\prime }\left(x\right)=3x^2-6x+3=3(x-1{)}^2$
Integrating again, we get $f^{\prime }\left(x\right)=(x-1{)}^3+D$
The curve passes through $(2,1)$
$\Rightarrow 1=(2-1)3+D\Rightarrow D=0$
$\therefore f\left(x\right)=(x-1{)}^3$