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The Weibull distribution starts at $x=0$ and has a single mode when its shape parameter is larger than one. See the green curve below for an example. Also, when the shape parameter is greater than one, the hazard rate is monotonically increasing. My interpretation of the Hazard rate of a distribution is - "how rapidly are the events that the distribution is modelling hitting you". If this is fair, then there seems to be an intuitive problem. For the distribution described by the green curve below, I would expect events to hit me at a low rate at first, then at a substantially higher rate as I came close to its mode and then I would expect them to hit me at a lower rate when I crossed the mode and entered the tail. But, we can see mathematically that the hazard rate is in fact, monotonically increasing and doesn't rise and then fall. What am I missing?

Weibull probability density function

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    Important question to consider: what is the random variable described by the Weibull distribution? The time to what? – 2017-02-20
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    Lets say its the time until a machine fails. – 2017-02-20
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    Okay, good. Now what would you expect the distribution of that to look like if the hazard rate were monotonically increasing? (Your intuition for the hazard rate is correct.) – 2017-02-20
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    Hmm.. I'll need to think about that. I feel that if it rises too fast, it might be impossible to even construct a distribution. That's a good pointer. – 2017-02-20
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    The distribution won't be monotonically increasing. At some point in time it will become very likely that it has *already* failed and the pdf of the failure time will become very small. – 2017-02-20
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    I see, thanks. You could have added that as an answer, but this works too :) – 2017-02-20

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Assume the density function for the lifetime distribution is Weibull distributed: $f(t;\lambda, k) = k \lambda^k t^{k-1} \exp\{-(\lambda t)^k\}$.
From here on, I will drop $\lambda$ and $k$ as function inputs for notional convenience.

Notice $\frac{\partial}{\partial t}\exp\{-(\lambda t)^k\} = - \exp\{-(\lambda t)^k\} \frac{\partial}{\partial t}(\lambda t)^k = -\lambda k (\lambda t)^{k-1}\exp\{-(\lambda t)^k\} = - f(t)$ using the chain rule.

From the density function, we derive the survival function using the previous result as follows: $S(t) := \int_{t}^{\infty } f(u) du= \Big|-\exp\{-(\lambda u)^k\}\Big|_t^\infty = 0 - (-\exp\{-(\lambda t)^k\} = \exp\{-(\lambda t)^k\}$.

We finally derive the hazard function using the survival and density functions as follows: $h(t) := \frac{f(t)}{S(t)} = k \lambda^k t^{k-1}.$

To explore how the hazard function changes over time, we shall analyze the partial derivative with respect to time: $\frac{\partial}{\partial t} h(t) = (k-1) k \lambda^k t^{k-2} $.

Since $\lambda, k$ are positive and $t$ is non-negative, we can see that:
For $k > 1$: the derivative is positive, so the hazard function is monotonically increasing.
For $ k < 1$: the derivative is negative, so the hazard function is monotonically decreasing.
For $k=1$: the derivative is zero, so the hazard function is constant.

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    Thanks. This is certainly useful, but I was looking for a more intuitive explanation, like the one provided by @spaceisdarkgreen in the comments.. – 2018-01-28
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    Oh yes, I see now that this doesn't really answer your original question and you said you understand the math already. Regardless, I'll leave this up for future readers. – 2018-01-28