Question

lf $f$ : $R\to R$ is a function such that $f\left(x\right)=x^3+x^2{f}^{{}^{\prime }}\left(1\right)+x{f}^{{}^{\prime \prime }}\left(2\right)+f^{{}^{\prime \prime \prime }}\left(3\right)$ $fo{r}_X\in R$ then $f\left(2\right)=$

• A. $2$
• B. $-2$
• C. $f\left(1\right)$
• D. $f\left(0\right)$

Answer 1

The correct option is B $-2$

$f^{\prime }\left(x\right)=3x^2+2x{f}^{{}^{\prime }}\left(1\right)+f^{{}^{\prime \prime }}\left(2\right)$
$f^{{}^{\prime }}\left(1\right)=3+2f^{{}^{\prime }}\left(1\right)+f^{{}^{\prime \prime }}\left(2\right)$
$f^{{}^{\prime \prime }}\left(2\right)+f^{{}^{\prime }}\left(1\right)+3=0----1$
$f^{{}^{\prime \prime }}\left(x\right)=6x+2f^{{}^{\prime }}\left(1\right)$
$f^{{}^{\prime \prime }}\left(2\right)=12+2f^{{}^{\prime }}\left(1\right)----2$
$f^{{}^{\prime }}\left(1\right)=-5$
$f^{{}^{\prime \prime }}\left(2\right)=2$
$f^{{}^{\prime \prime \prime }}\left(3\right)=6$
$f\left(2\right)=8+4(-5)+(2\times 2)+6$
$=-2$