Respuesta :

[tex]y=4e^x+x\implies y'=4e^x+1[/tex]

At the point (0,1), the derivative has the value of [tex]y'=4e^0+1=5[/tex], so this is the slope of the tangent line. The equation for the line is then

[tex]y-4=5(x-0)\implies y=5x+4[/tex]