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Analysis Homework #4 Solutions 1. Find the values of x for which 6x3 + x < 5x2 . • To ﬁnd the values of x for which f (x) = 6x3 − 5x2 + x is negative, we write f (x) = x(6x2 − 5x + 1) = x(2x − 1)(3x − 1) and then worry about the signs of each of the factors separately. x x 2x − 1 3x − 1 f (x) 0 − − − − 1/3 + − − + 1/2 + − + − + + + + According to the table, f (x) is negative when x < 0 and also when 1/3 < x < 1/2. 2. Compute each of the following limits: L = lim x→+∞ 1 + 2x − x2 , 3 − 2x + x2 M = lim x→−∞ x3 − 2x + 1 . x4 + 2x − 3 • Dividing both the numerator and the denominator by x2 , we easily get L = lim x→+∞ 1 + 2x − x2 1/x2 + 2/x − 1 0+0−1 = lim = = −1. 2 2 x→+∞ 3/x − 2/x + 1 3 − 2x + x 0−0+1 Dividing both the numerator and the denominator by x4 , we similarly get M = lim x→−∞ x3 − 2x + 1 1/x − 2/x3 + 1/x4 0−0+0 = lim = = 0. 4 3 4 x + 2x − 3 x→−∞ 1 + 2/x − 3/x 1+0−0 3. Use the definition of the derivative to compute f ′ (y) when f (x) = ax3 + bx2 + cx + d for some constants a, b, c, d ∈ R. • In this case, we have f (x) − f (y) a(x3 − y 3 ) + b(x2 − y 2 ) + c(x − y) = x−y x−y and we can simplify the expression on the right hand side to get f (x) − f (y) = a(x2 + xy + y 2 ) + b(x + y) + c. x−y Using the deﬁnition of the derivative, we now ﬁnd [ ] f (x) − f (y) = lim a(x2 + xy + y 2 ) + b(x + y) + c x→y x→y x−y f ′ (y) = lim and polynomials are known to be continuous, so f ′ (y) = a(y 2 + y 2 + y 2 ) + b(y + y) + c = 3ay 2 + 2by + c. 4. Find the min and max values of f (x) = 3x4 − 16x3 + 18x2 over [−1, 1]. • Since f is continuous on a closed interval, it suﬃces to check the endpoints, the points at which f ′ does not exist and the points at which f ′ is equal to zero. In this case, f ′ (x) = 12x3 − 48x2 + 36x = 12x(x2 − 4x + 3) = 12x(x − 1)(x − 3) so the only points at which the min/max values may occur are the points x = −1, x = 1, x = 0, x = 3. We exclude the point x = 3, which does not lie in the given interval, and we note that f (−1) = 37, f (1) = 5, f (0) = 0. Thus, the minimum value is f (0) = 0 and the maximum value is f (−1) = 37.