On real functions with the property that for each $\varepsilon>0,$ every interval contains two points whose secant line is $>\varepsilon.$

Let $f:\mathbb{R}\to\mathbb{R}$ have the following property:

$$\forall\varepsilon>0,\text{ no matter how large },\forall a<b,\ \exists a<x_1<x_2<b\text{ such that }\left\lvert\frac{f(x_2)-f(x_1)}{x_2-x_1}\right\rvert>\varepsilon.$$

In other words, for every $\varepsilon>0,$ every interval contains two points whose (absolute value of) gradient of secant line is $>\varepsilon.$

Obviously $f$ can be discontinuous - in fact, every "everywhere discontinuous" function satisfies the above property. I am not sure if $f$ can be continuous, but perhaps the Weierstrass function is one example (but I would like to confirm or deny this). But I think the function cannot be differentiable. So my questions are:

Can such a function be continuous? If yes, can it be differentiable?