$\text{ If the function }f\left(x\right)=x^4+b{x}^2+8x+1\text{ has a horizontal tangent and a point }x\text{ for }{f}^{\prime \prime }\left(x\right)=0$  $\text{ then the value of b is equal to }$

$f^{\mathrm{\prime }}\left(x\right)=0\text{ and }{f}^{\prime \prime }\left(x\right)=0\text{ for same }x=x_1\text{ (Say) }$

$\begin{array}{l}\Rightarrow f^{\mathrm{\prime }}\left(x\right)=4x^3+2bx+8\Rightarrow f^{\prime \prime }\left(x\right)=12x^2+2b\\ \Rightarrow f^{\mathrm{\prime }}\left(x_1\right)=2\left(2x_1^3+b{x}_1+4\right)=0\to \left(1\right)\\ \Rightarrow f^{\prime \prime }\left(x_1\right)=2\left(6x_1^2+b\right)=0\to \left(2\right)\\ \Rightarrow b=-6x_1^2\\ \text{ Substitute in eqn }->1\\ 2x_1^3+\left(-6x_1^3\right)+4=0\Rightarrow 4x_1^3=4\Rightarrow x_1^3=1\\ \Rightarrow x_1=1\Rightarrow b=-6\end{array}$