Let f: R rightarrow R be a function such that | vedclass

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(B) $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6$ $	herefore f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2)$ ...$(i)$ $	herefore

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-$2$

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(B) $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6$ $	herefore f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2)$ ...$(i)$ $	herefore f^{\prime \prime}(x)=6 x+2 f^{\prime}(1)$ ...(ii) Substituting $x=1$ in $(i)$,we get $f^{\prime}(1)=3(1)^2+2(1) f^{\prime}(1)+f^{\prime \prime}(2)$ $\Rightarrow f^{\prime}(1)+f^{\prime \prime}(2)=-3$ ...(iii) Substituting $x=2$ in (ii),we get $f^{\prime \prime}(2)=6(2)+2 f^{\prime}(1)$ $\Rightarrow f^{\prime \prime}(2)=12+2 f^{\prime}(1)$ ...(iv) From (iii) and (iv),we get $f^{\prime}(1)+12+2 f^{\prime}(1)=-3$ $\Rightarrow 3 f^{\prime}(1)=-15$ $\Rightarrow f^{\prime}(1)=-5$ From (iii),$-5+f^{\prime \prime}(2)=-3$ $\Rightarrow f^{\prime \prime}(2)=2$ $	herefore f(2) = 2^3+2^2(-5)+2(2)+6$ $= 8-20+4+6$ $= -2$

Let $f: R \rightarrow R$ be a function such that $f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+6, x \in R$,then $f(2)$ equals

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