Exam 1. (2x + 1) 2 9. lim. (rearranging) (x 1 implies x 1, thus x 1 0
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1 Department of Mathematical Sciences Instructor: Daiva Pucinskaite Calculus I January 28, 2016 Name: Exam 1 1. Evaluate the it x 1 (2x + 1) 2 9. x 1 (2x + 1) 2 9 4x 2 + 4x = 4x 2 + 4x 8 = 4(x 1)(x + 2) = = x 1 4(x + 2) = 4(1 + 2) = 12 (rearranging) (x 1 implies x 1, thus x 1 0 divide the numerator and denominator by x 1) (p(x) = 4(x + 2) is a polynomial thus x 1 p(x) = p(1))
2 2. Finding its from a graph. Use the graph of f in the figure to find the following values or state that they do not exist. If a it does not exist, explain why. a. f(3) = 3 b. f(x) = 3 x 3 + d. x 3 f(x) does not exist e. x 9 + f(x) = 7 c. f(x) = 1 x 3 f. f(x) = 3 x 9 g. f(9) = 5 h. f(x) = 7 i. f(x) = 7 x 13 + x 13 j. x 13 f(x) = 7 k. f(5) = 5 l. x 5 f(x) = 5 d. f(x) does not exist, because f(x) f(x). x 3 x 3 } + x 3 {{}} {{}
3 3. Limits with a parameter. Let f(x) = x2 6x + 8 x a For p(x) = x 2 6x + 8 = (x 2)(x 4) and q(x) = x a we consider the rational function f(x) = p(x) q(x) = x2 6x + 8 (x 2)(x 4) =. x a x a a. For what values of a, if any, does f(x) equal a finite number? Since q(x) = (x a) = 0, we have that p(x) f(x) = q(x) = x 2 6x + 8 (x 2)(x 4) = x a x a is a finite number, if numerator and denominator can be divided by g(x) = x a. This is possibe if (x 2)(x 4) (x 2)(x 4) x 2 a = 2, because = x a + x 2 or (x 2)(x 4) (x 2)(x 4) x 4 a = 4, because = x a x 4 + x 4 b. For what values of a, if any, does f(x) =? = = x 4 +(x 4) = 2, +(x 2) = 2 x a + implies a < x, we have that g(x) = x a > 0 for all x near a from the right. Moreover g(a) = a a = 0. Therefore if p(a) = (a 2)(a 4) = L > 0. p(x) f(x) = q(x) = (x 2)(x 4) = x a If a > 4, then p(a) = (a 2) (a 4) > 0, >0 >0 If a < 2, then p(a) = (a 2) (a 4) > 0. <0 <0 >0 (x 2)(x 4) For any a R such that a > 4, or a < 2 we have f(x) = =. x }{{ a } x a 0 and x a>0 c. For what values of a, if any, does f(x) =? Since g(x) = x a > 0 for all x near a from the right and g(a) = a a = 0 we have
4 p(x) f(x) = q(x) = (x 2)(x 4) = x a if p(a) = (a 2)(a 4) = L < 0. This is the case when (a 2) > 0 and (a 4) < 0, i.e. a > 2 and a < 4 or (a 2) < 0 and (a 4) > 0, but there are no real numbers with a < 2 and a > 4. <0 (x 2)(x 4) For any a R such that 2 < a < 4 we have f(x) = =. x }{{ a } x a 0 x a>0
5 4. Consider the function f given by Recall: x ± a. Analyze f(x) and f(x) = x2 16 x(x 4) c n x n + c n 1 x n c 1 x + c 0 d m x m + d m 1 x m d 1 x + d 0 = x ± c n x n d m x m. f(x), and then identify any horizontal asymptotes. x The line y = L is a horizontal asymptote of f if f(x) = L or f(x) = L. x x 2 16 f(x) = x(x 4) = x 2 16 x 2 4x = x 2 x 2 x f(x) = x 2 16 x x 2 4x = x 2 x x 2 x x 0 = x 1 = 1 x 0 = 1 = 1 The line y = 1 is the only horizontal asymptote of f, because f(x) = 1 as well as x f(x) = 1 b. Find the vertical asymptotes. Analyze f(x) and f(x) for each vertical asymptote x = a. x a The line x = a is vertical asymtote, if f(x) = ±, or We have f(x) = ± x a x 2 16 f(x) = x a ± x a ± x(x 4) = (x 4)(x + 4) = ± if x a ± x(x 4) x(x 4) is a 16) 0 and x(x 4) polynomial = a(a 4) = 0, i.e. a = 0 or a = 4 x a ±(x2 x a ± x 2 16 is a polynomial If a = 0, then 16) = = 16 0, and x(x 4) = 0. x 0 ±(x2 x 0 ± Therefore, the line x = 0 is a vertical asymptote of f. We have f(x) = x 0 x 0 16<0 x 2 16 x(x 4) x 0 thus x<0 x(x 4)>0 =, and f(x) = x 0 + x 0 + (x 4)(x + 4) x 4 (x + 4) If a = 4, then f(x) = = x a ± x 4 ± x(x 4) x 4 ± x The line x = 4 isn t a vertical asymptote of f. x 4 ± thus x 0 16<0 x 2 16 x(x 4) x 0 + thus 0<x<4 x(x 4)<0 = = = 2 ±.
6 5. An unknown constant. Let x 2 + x if x < 2 f(x) = b if x = 2 3x + 7 if x > 2. a. Determine the value of b for which f is continuous from the left at 2. A function f is continuous from the left at x = a, if f(x) = f(a) x a x 2 implies x < 2, thus in this case we have f(x) = x 2 + x. Since f(x) = + x) = = 6 and f(2) = b, we have (x2 for b = 6. f(x) } {{} 6 = f(2) b b. Determine the value of b for which f is continuous fro m the right at 2. A function f is continuous from the right at x = a, if f(x) = f(a) x 2 + implies x > 2, thus in this case we have f(x) = 3x + 7. Since f(x) = + +(3x + 7) = = 13 and f(2) = b, we have for b = 13. f(x) } + {{} 13 = f(2) b c. Is there a value of b for which f is continuous at 2? Explain. A function f is continuous at x = a, if x a f(x) = f(a) Since f(x) } {{} 6 + f(x) 13 f(x) does not exist. There is no b with f(x) = f(2) = b.
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