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

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To find the critical points of ( h(x) = e^{-x} + kx ), we need to find where its derivative equals zero:

[ h'(x) = -e^{-x} + k ]

Setting ( h'(x) = 0 ), we have:

[ -e^{-x} + k = 0 ]

[ e^{-x} = k ]

[ e^x = \frac{1}{k} ]

[ x = \ln\left(\frac{1}{k}\right) ]

This gives us a critical point at ( x = \ln\left(\frac{1}{k}\right) ). For this to be a valid critical point, ( \ln\left(\frac{1}{k}\right) ) must be in the domain of ( h(x) ), which means ( k ) must be positive. So, the critical point exists for ( k > 0 ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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