Question

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?

Answer 1

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 `kx` 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\leq 0`, the derivative has no zeros, and so the function has no critical points.