What are the solution?

Explanation. This problem is more challenging because we have more unknowns. However, be brave intrepid mathematician. To find a and 6 that make f continuous at z = 1, we need to find a and b such that imsey=1 Let us begin with item (a) of the continuity checklist at z 1. f(1) a +6 + 2 exists for all values of a and 6, so this item is satisfied. We now check item (b) of the checklist by verifying that lim f(r) exists. In order to calculate lims 1 f(r), we have to compute two one-sided limits (since the expression for f(z) ifr 1). Looking at the limit from the left, | B) B we have Looking at the limit from the right, we have tim, f(z) = lim, (az' + br +2) Hence, for the limit lim, ,, f(z) to exist, we must have that Item (c) is then satisfied at the point z = 1 whenever a+b = 3. To find a and 6 that make f is continuous at z = 3, we need to find a and 6 such that jim f(z) = (3). Let us return to the continuity checklist at the point z = 3. For item (a) notice that f(3) = 18 + a b, which exists for all values of @ and b. 'To verify item (b) we must check that lim, ys f(z) exists. Again, the formula for f(z) changes around z = 3 so we check the one-sided limits. From the left: lim = lim (az? + br +2 rn oats) =a-94b-342. Looking at the limit from the right, we have lim, f(z) = lim (62 +a 8)