1. ## error bounds

Is it the the maximum on the closed interval [a,b] not the largest local maximum, correct? (unless that largest local maximum is also the largest value on the closed interval)

and what if the function ends up having a vertical asymptote on the interval? such has 6/x on [0,3]

2. ## Re: error bounds

Originally Posted by Jonroberts74
Another quick question about error bounds.
Pretty sure that both of us just plotted f'' and eyeballed that the maximum was at x=0 but more formally

if you had taken the 3rd derivative you would find that it is non-zero on [0,3].

Thus the maximum has to be at a boundary point of your interval.

Clearly |f''(3)| < |f''(0)| so f''(0) must be the maxima on [0,3]

most likely a vertical asymptote within your limits of integration mean that your function isn't integrable on this interval.

3. ## Re: error bounds

this is for k_{4} which is the fourth derivative is what I am looking at now the fourth derivative gives me 6/x. so how can I find the error bounds if I cant put a maximum on that interval?