Let f(x)=x^3+x^2f^'(1)+xf^''(2)+f^'''(3),x R.

Question

Let $f(x)={x}^{3}+{x}^{2}{f}^{'}(1)+x{f}^{''}(2)+{f}^{'''}(3),x\in R$. Then ${f}^{'}(10)$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 202

Accepted Answer

$\begin{gathered} f(x)={x}^{3}+{x}^{2}\cdot {f}^{'}(1)+x\cdot {f}^{''}(2)+{f}^{'''}(3) \ {f}^{'}(x)=3{x}^{2}+2x{f}^{'}(1)+{f}^{''}(2) \ {f}^{''}(x)=6x+2{f}^{'}(1) \ {f}^{'''}(x)=6 \ {f}^{'}(1)=-5,{f}^{''}(2)=2,{f}^{'''}(3)=6 \ f(x)={x}^{3}+{x}^{2}\cdot (-5)+x\cdot (2)+6 \ {f}^{'}(x)=3{x}^{2}-10x+2 \ {f}^{'}(10)=300-100+2=202 \end{gathered}$

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