Question

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

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

Answer 1

Answer: (B)

$f″ \left(x\right) = 6\left(x - 1\right) ⇒ f′ \left(x\right) = 3\left(x - 1\right)^{2 }+ c$
and $f′ \left(2\right) = 3 ⇒ c = 0$
$⇒ f \left(x\right) = \left(x - 1\right)^{3} + k$ and $f \left(2\right) = 1 ⇒ k = 0$
$⇒ f \left(x\right) = \left(x - 1\right)^{3}.$