# How do you find values of k for which there are no critical points if h(x)=e^(-x)+kx where k is any constant?

Mar 10, 2015

First of all, a critical point of a function $h$ is defined as a point in which the derivative $h '$ equals zero. In your case, the derivative is
$- {e}^{- x} + k$
in fact, applying the chain rule, the derivative of ${e}^{- x}$ is ${e}^{- x}$ times the derivative of $- x$, which is $- 1$; while the derivative of $k x$ is of course $k$
The question is: for which values of $x$ we have
$- {e}^{- x} + k = 0$
easy manipulations bring us to the equivalent request
${e}^{- x} = k$
Since ${e}^{- x}$ is always (strictly) positive, this equation has solutions only if $k$ is positive.

So, the final answer is: if $k \setminus \le q 0$, the derivative has no zeros, and so the function has no critical points.