quiz10Asol

# quiz10Asol - Solutions to Quiz 10A...

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Unformatted text preview: Solutions to Quiz 10A www.math.uﬂ.edu/˜harringt April 6, 2008 1. Find the limit. Use l’Hosptial’s Rule where appropriate. If there is a more elementary method, consider using it. (a) limx→0 1−cos(x) x2 lim x→0 1 − cos(x) L sin x L cos x cos(0) 1 = lim = lim = = x→0 2x x→0 x2 2 2 2 (b) limx→∞ x1/x 1 ln x L 1 = lim x = lim =0 x→∞ x x→∞ 1 x→∞ x ln( lim x1/x ) = lim ln(x1/x ) = lim x→∞ x→∞ lim x1/x = e0 = 1 x→∞ (c) limx→π sin(x) 1−cos(x) lim x→π sin(π ) 0 sin(x) = = =0 1 − cos(x) 1 − cos(π ) 1 − (−1) Notice, here we did not use l’Hospital’s rule at all. 2. Let f (x) = −x3 + 9x2 − 15x + 2. Using interval notation, determine where f is concave down. Also, ﬁnd the absolute maximum. First we will ﬁnd when the f is concave down. So we look at the second derivative. f (x) = −3x2 + 18x − 15 = −3(x − 5)(x − 1) f (x) = −6x + 18 (1) (2) The function f is concave down when f (x) < 0, so we have −6x + 18 < 0 implies that x > 3. So f is concave down on (3, ∞). What about the absolute maximum? Well, notice that f is a polynomial of odd degree, so f does not have an absolute maximum! Most of my students found the relative maximum (instead of the absolute maximum), so if you were curious, here is how to ﬁnd the relative maximum. Notice that f (x) = −3x2 + 18x − 15 = −3(x − 5)(x − 1), so our critical numbers are x = 5 and x = 1. Since, f (1) > 0, then the point at x = 1 is a local minimum and f (5) < 0, then the point at x = 5 is a local maximum. The graph below is of f . Notice it has a local max, but as x → −∞, f goes oﬀ to inﬁnity. Figure 1: f (x) = −x3 + 9x2 − 15x + 2 1 ...
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