Let h(x) = e^(-x) + kx, where k is any constant. For what value(s) of k does h have critical points?

It has critical points only for $k > 0$

First, let's compute the first derivative of $h \left(x\right)$.

${h}^{p r i m e} \left(x\right) = \frac{d}{\mathrm{dx}} \left[{e}^{- x} + k x\right] = \frac{d}{\mathrm{dx}} \left[{e}^{- x}\right] + \frac{d}{\mathrm{dx}} \left[k x\right] = - {e}^{- x} + k$

Now, for ${x}_{0}$ to be a critical point of $h$, it must obey the condition ${h}^{p r i m e} \left({x}_{0}\right) = 0$, or:

${h}^{p r i m e} \left({x}_{0}\right) = - {e}^{- {x}_{0}} + k = 0 \iff {e}^{- {x}_{0}} = k \iff - {x}_{0} = \ln \left(k\right) \iff$
$\iff {x}_{0} = - \ln \left(k\right)$

Now, the natural logarithm of $k$ is only defined for $k > 0$, so, $h \left(x\right)$ only has critical points for values of $k > 0$.