Point of intersection $g(x)=e^{-x^2/c}$ and curve $f(x)=x^2+4x+4$.But using derivative

Question

I am trying to find (point of touch) of two function

1.$g(x)=e^{-x^2/c}$ (gaussian)

2.$f(x)=x^2+4x+4$ (quadratic)

I approached by equating tangents of both equation to equal.Thus I equated following equation..::..

$ 2x+4=-2x^3/c\cdot e^{-x^2/c}$

But i caught up with lograthmic equation which goes like this..

$ \ln((2x^3/c)+2x+4)=-x^3/c$

I dont know how shall i proceed from here.. Thank You

Answer 1

Hint: the equations

$f(x)=g(x)$ and $f'(x)=g'(x)$

lead to

$2c(x+2)=-2x(x+2)^2$.