I'm not a big fan of that picture. We're trying to
compute f'(400), judging by the text (scroll down a bit on the first
page). What the image suggests, afaict, is that f'(400) that can be
computed as the limit [f(x+h) - f(x-h)]/2h with x=400 and h -> 0 (in
the picture, we see the case h=80). At least the graphic appears to
connect the two somehow.
Nothing to that effect is mentioned in the text... just the cookbook
recipe 'at^n becomes ant^{n-1}'. So why is there even a triangle here?
Two things are fishy here. The limit and the triangle.
The limit: Taking the aforementioned limit for the derivative is
rather weird. If the function is differentiable, you'll get the right
result, namely f'(x). If you consider the limit for the function x ->
|x| at x=0 e.g., you'll get a limit as well, even though that function
is not differentiable.
The triangle: I would expect it to connect the three points
[x-h,f(x-h)], [x+h,f(x-h)], [x+h,f(x+h)]. Instead, we get three points
whose X-coordinates match the ones I mentioned with Y-coordinates that
are chosen to have the slope of the triangle match the slope of the
function. Why? (If this were about the mean-value theorem... but it's not!)
So what I'm left with is a plot with a triangle, no text that explains
any of that, and nothing I can make of it myself.
http://minireference.com/miniref/lib/tpl/miniref/landings/im...
I'm not a big fan of that picture. We're trying to compute f'(400), judging by the text (scroll down a bit on the first page). What the image suggests, afaict, is that f'(400) that can be computed as the limit [f(x+h) - f(x-h)]/2h with x=400 and h -> 0 (in the picture, we see the case h=80). At least the graphic appears to connect the two somehow.
Nothing to that effect is mentioned in the text... just the cookbook recipe 'at^n becomes ant^{n-1}'. So why is there even a triangle here? Two things are fishy here. The limit and the triangle.
The limit: Taking the aforementioned limit for the derivative is rather weird. If the function is differentiable, you'll get the right result, namely f'(x). If you consider the limit for the function x -> |x| at x=0 e.g., you'll get a limit as well, even though that function is not differentiable.
The triangle: I would expect it to connect the three points [x-h,f(x-h)], [x+h,f(x-h)], [x+h,f(x+h)]. Instead, we get three points whose X-coordinates match the ones I mentioned with Y-coordinates that are chosen to have the slope of the triangle match the slope of the function. Why? (If this were about the mean-value theorem... but it's not!)
So what I'm left with is a plot with a triangle, no text that explains any of that, and nothing I can make of it myself.