$(25pts.)$ The equation $e^x-4x-1=0$ can $be$ written $in$ the form $x=g(x)in$ the following ways,

$x_{n+1}=e^{x_n}-3x_n-1,x_{n+1}=ln(4x_n+1),x_{n+1}=\frac{e^{x_n}-1}{4}$

Which sequence converges $to$ the unique fixed point $of$ the equation $e^x-4x-1=0in1,2.$

Use the convergent sequence $to$ find $x_{2}by$ taking $x_0=1\mathrm{.}1.$