Question

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

Answer 1

It has critical points only for `k>0`

First, let's compute the first derivative of `h(x)`.

`h^(prime) (x) = d/(dx)[e^(-x)+kx] = d/(dx)[e^(-x)] + d/(dx)[kx] = - e^(-x) +k`

Now, for `x_0` to be a critical point of `h`, it must obey the condition `h^(prime)(x_0) = 0`, or:

`h^(prime) (x_0) = -e^(-x_0)+k = 0 <=> e^(-x_0) = k <=> -x_0 = ln(k) <=>`
`<=> x_0 = -ln(k)`

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