Surprisingly, I could find no answers for this on either YouTube or the internet. I'm looking for the average rate of change on the interval $[-1,3]$ of the function: $$f(t)=6t^2+\frac{4}{t^2}$$ Now, despite the function being non-differentiable at $0$, we can still easily take the derivative: $$f'(t)=12t-\frac{8}{t^3}$$ And I can easily find the average value by hand: $$ave_{f'}=\frac{1}{4}\int_{-1}^3 12t-\frac{8}{t^3} \,dt=\frac{6t^2+\frac{4}{t^2}\Big|_{-1}^3}{4}=\frac{100}{9}$$ However, I am afraid that this method is improper due to the discontinuity. If I put that same integral into either Desmos or Symbolab, it complains and says that the integral diverges. I can think of two alternate methods to solve:
$$\frac{1}{4}\left[\lim_{a\to0^-} \int_{-1}^a 12t-\frac{8}{t^3} \,dt+\lim_{a\to0^+}\int_{a}^3 12t-\frac{8}{t^3} \,dt\right]$$
Or, using the fact that the function is even (reflection over y-axis) means that the derivatives from $-1$ to $0$ and from $0$ to $1$ will cancel each other out, and we can simply take the average value from $1$ to $3$: $$ave_{f'}=\frac{1}{4}\left[0+\int_{1}^3 12t-\frac{8}{t^3} \,dt\right]=\frac{6t^2+\frac{4}{t^2}\Big|_{1}^3}{4}=\frac{100}{9}$$ Which is the proper method to explain to my student? Thanks!