The golden-section search algorithm involves dividing the search interval into two sub-intervals with a golden ratio (approximately 0.618) and evaluating the function at the endpoints of these sub-intervals. The sub-interval that does not contain the maximum is discarded, and the search continues with the remaining sub-interval.

In the first iteration, we have:

xL = 0, xU = 2 d = 0.618(2-0) = 1.236 x1 = xU - d = 0.764 x2 = xL + d = 1.236 f(x1) = -1.5(0.764)^6 - 2(0.764)^4 + 12(0.764) = 4.267 f(x2) = -1.5(1.236)^6 - 2(1.236)^4 + 12(1.236) = -3.267 New interval: [0.764, 2]

In the second iteration, we have:

xL = 0.764, xU = 2 d = 0.618(2-0.764) = 1.236(0.618) = 0.764 x1 = xU - d = 1.236 x2 = xL + d = 1.528 f(x1) = -1.5(1.236)^6 - 2(1.236)^4 + 12(1.236) = -3.267 f(x2) = -1.5(1.528)^6 - 2(1.528)^4 + 12(1.528) = -2.267 New interval: [0.764, 1.528]

In the third iteration, we have:

xL = 0.764, xU = 1.528 d = 0.618(1.528-0.764) = 0.764(0.618) = 0.472 x1 = xU - d = 1.056 x2 = xL + d = 1.292 f(x1) = -1.5(1.056)^6 - 2(1.056)^4 + 12(1.056) = 0.267 f(x2) = -1.5(1.292)^6 - 2(1.292)^4 + 12(1.292) = -1.267 New interval: [1.056, 1.528]

Therefore, the value of x1 at the end of the third iteration is 1.056, which corresponds to the left endpoint of the final interval. The answer is (c) x1=1.7448.