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

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Answer 1 · 2015-02-21T07:07:57

It has critical points only for $k > 0$ $k>0$

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

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

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 $k>0$ .

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