Finding value of k so two curves are tangent

[songoku]

Homework Statement

Find the smallest value of k (where k ≥ 1) so that y = sin x and $y=ke^{-x}$ are tangent for x ≥ 0

Relevant Equations

Derivative

I tried to equate the derivative of the two equations:
$$\cos x=-ke^{-k}$$

Then how to continue? Is this question can be solved?

Thanks

 

[Delta2]

In order for two curves $y=f(x),y=g(x)$ to be tangent at a point $x_0$ two conditions should hold:

1. $f(x_0)=g(x_0)$
2. $f'(x_0)=g'(x_0)$

So put these two conditions into work for the specific $f(x)=\sin x$ and $g(x)=ke^{-x}$. You ll get a system of two equations with two unknowns $x_0$ and $k$. It will turn out that you will have many possible values for $x_0$ (trigonometric equation involved) and from the corresponding values for $k(x_0)$ there will be a minimum.

 

[songoku]

Thank you very much Delta2