Test 1: typical mistakes Problem 2: naive interval computations should be applied to the original function, NOT to its transformed form; when you make a transformation, the function remains the same but the result of applying naive interval computation changes. For example, for f(x)=2x-x, if you apply it to x=[0,1], you get 2[0,1]-[0,1]=[0,1]-[0,1]=[-1,2], but for the equivalent form f(x)=x, we get a different answer [0,1]. Both are correct in the sense that both are enclosures Problem 3: I explicitly asked to apply naive interval computations only when the function is NOT monotonic, and it happens to be monotonic, so you should apply the fact that for a monotonic function on the interval [a,b], the range is [f(a),f(b)] (when increasing) or [f(b),f(a)] (if decreasing) Problem 4: * On centered form with checking monotonicity first, if you checked and the function is monotonic, you get its range, there is no need to apply centered form * Once you have ranges for both subintervals, you need to take the union of these ranges to get the range on the original interval Problem 5: for x1 on [a1,b1], either x1 = a1, or x1 = b1, or the partial derivative with respect to x1 is equal to 0; similar, for x2, we have 3 similar options, so we need to consider 9 possible cases, they nicely fit into the table, when 3 cases for x1 are rows and 3 cases for x2 are columns Problem 7: the task was NOT to find the range, but to LOCATE the maximum, i.e., to find the values x at which maximum can be attained, we did that by dismissing some subintervals; in this particular case, since the function is increasing on each subinterval, we dismiss all subinterval except for its right-end point; out of two right-end points, the first one is smaller than maximum so far, so we end up with a single location x = 0 Problem 8: you bisect ONLY when after the whole cycle, i.e., after applying ALL the rules, you get the same intervals for all the unknown. Bisection doubles the time, so if you apply it every time for n steps, you need 2^n steps - which makes the algorithm not feasible. DO NOT bisect over ALL intervals - this again leads to 2^n cases and is, thus, in general, not feasible, you only bisect over ONE of the variables