Test 2: typical mistakes

Problem 1-2:

1) some of you confuse linearization with centered form,
linearization is an approximate fast method, in which we do not use
interval computations at all

2) the task is to compute the ranges, so if you only computed Δ,
you are not done yet, you still need to compute the range
[y − Δ, y + Δ]

3) several mistakes when computing the range of the function using
calculus (by the way, you were supposed to compute by using
calculus); for a function f9x) of one variable, the range on an
interval [a, b] is computed as follows: we compute f(a), f(b), and
we compute the values f(x) for all the point x from the interval
[a,b] at which f'(x) = 0; then the minimum of all these values is
the lower endpoint of the range, the maximum is the upper endpoint

in this case, the value x wt which f'(x) = 0 is outside the
intervals, but even if it was inside, you should take the value
f(x) when computing min and max, NOT the value x itself as some of
you did

Problem 3: \/ ("or") is max and & (and) is min, not vice versa

Problem 4: the only way to solve an equation with absolute value
is to consider two cases: when the absolute value is non-negative,
and when it is negative

Problem 7: bisection is a very simple algorithm, you compute the
value at midpoint, and depending on the sign of this value, divide
the interval in half; there is no evaluation of derivatives here

Problem 9: some of you forgot to answer the very first question: s
there an alternative which is guaranteed to be optimal?

Problem 10: for each example, you need to show that this improves
or worsens the results of naive interval computations, and the
only way to do it is by providing a numerical example; also, both
examples should be different from what I taught

Problem 11: the goal of interval arithmetic is to produce
enclosures, that guarantee to contain all possible values; to get
a guaranteed bound for an interval, we need to round the left
endpoint to the left, and the right endpoint to the right; so
[0.019, 0.021] should be rounded to [0.01,0.03].