Certain graphs, specifically those that are differentiable, have a property called local linearity. This means that if you zoom in (using the same zoom factor in both directions) on a point on the graph, the graph eventually appears to be a straight line whose slope if the same as the slope (derivative) of the tangent line at that point.
Now we are a little ahead of ourselves here since we haven’t mentioned tangent lines and derivatives yet. But local linearity is the graphical manifestation of differentiability. Functions that are differentiable at a point are locally linear there and functions that are locally linear are differentiable. In the next post we will see how to use the local linear idea to introduce the derivative. For now, we will look at some graphs that may or may not be locally linear. Are the graphs “smooth” everywhere?
The first function is locally linear at (0, 1) but doesn’t look it. Zoom-in several times and you will see that it is smooth and locally linear there eventually the graph looks like a horizontal line near (0, 1). This only looks like an absolute value graph because .
The second function appears locally linear at (1, 1), but is not. Zoom in a few times at you will see very strange things going on. (Hint: Use a graphing program or a calculator and enter x^3+((x – 1)^2)^(1/3)/7 as simplifying to a power of 2/3 may confuse the calculator.)
The moral is that you can never be sure just looking at a graph whether it is locally linear or not; you’re never sure if you have zoomed in enough.
Nevertheless, the local linearity concept is helpful in introducing the derivative and, when you can be sure the function is not locally linear, knowing the derivative does not exist.
Looking ahead: Take a look at the AP exam from 2005 AB 5. Here you were given a velocity graph “modeled by the piecewise- linear function defined by the graph” copied below.
Later they were asked if the Mean Value Theorem guaranteed a value of c in the interval [8, 20] such that v ‘(c) is equal to the average rate of change over the interval. The answer is “no”, the value does not exist because the function is not differentiable on the interval because it is not locally linear everywhere in the interval.