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].