MATH 121 Lecture : MATH130_BOYLE-M_SPRING2015_0101_MID_SOL_3
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So, f (x) = 0 at x = 2. Because f is di erentiable, the max and min values can only be assumed at inputs from {1, 2, 10}. Minimum value is f (2) = 4 8 ln(2) (by the rst derivative test, or by comparing values). Maximum value is f (10) = 100 8 ln(10) (this number is larger than f (1) = 1): (10 points) Find the equation of the tangent line to the curve 4e2x y2 = 0 at the point (0, 2). For (x, y) = (0, 2), we have y = 4e0/2 = 2, and an equation for that tangent line is y 2 = 2x : (15 points) Solution. p2(0) + 1 (1/2)+p2(1/2) + 1 (1/2)+p2(1) + 1 (1/2)+p2(3/2) + 1 (1/2) . (b) (5 pts) draw the graph of f and the rectangles corresponding to this.
