Let f(x) be a polynomial of degree 4 and $f\left(2\right)=-1,f\text{'}\left(2\right)=0,f\text{'}\text{'}\left(2\right)=2,f\text{'}\text{'}\text{'}\left(2\right)=-12,f\text{'}\text{'}\text{'}\text{'}\left(2\right)=24$ then find $f\text{'}\text{'}\left(1\right)=$

$\begin{array}{c}Letf\left(x\right)=a{x}^4+b{x}^3+c{x}^2+dx+e\Rightarrow f\left(2\right)=-1\Rightarrow 16a+8b+4c+2d+e=-1\to \left(1\right)\\ f\text{'}\left(x\right)=4a{x}^3+3b{x}^2+2cx+d\Rightarrow f\text{'}\left(2\right)=0\Rightarrow 32a+12b+4c+d=0\to \left(2\right)\\ f"\left(x\right)=12a{x}^2+6bx+2c\Rightarrow f"\left(2\right)=2\Rightarrow 48a+12b+2c=2\to \left(3\right)\\ f"\text{'}\left(x\right)=24ax+6b\Rightarrow f^{\text{'}\text{'}}\text{'}\left(2\right)=-12\Rightarrow 48a+6b=-12\to \left(4\right)\\ f""\left(x\right)=24a\Rightarrow f""\left(2\right)=24\Rightarrow 24a=24\Rightarrow a=1\\ \left(4\right)\Rightarrow 48+6b=-12\Rightarrow 6b=-60\Rightarrow b=-10\\ \left(3\right)\Rightarrow 48-120+2c=2\Rightarrow 2c=74\Rightarrow c=37\\ \left(2\right)\Rightarrow 32-120+148+d=0\Rightarrow d=-60\end{array}$

$\begin{array}{c}{f}^{\text{'}\text{'}}\left(1\right)=12\left(1\right){\left(1\right)}^2+6\left(-10\right)\left(1\right)+2\left(37\right)\\ =12-60+74\\ =26\end{array}$