Fall06-Test1-solutions

Fall06-Test1-solutions - Math 1205 -- Test 1 20 Sep 2006 f...

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Unformatted text preview: Math 1205 -- Test 1 20 Sep 2006 f Hx L 5 4 3 2 1 NAME: Explain all work! Use only methods developed in class. -6 -5 -4 -3 -2 -1 -1 -2 -3 -4 1 2 3 4 5 6 x 1. [16] a) Let f Hx L be defined by the graph given above. Determine the following limits, writing + or - as appropriate. If a limit does not exist, explain why. x - lim f Hx L lim x ض x H-1L- x 1- lim f Hx L f Hx L -4 -4 HDNEL 2 lim f Hx L lim x H-1L+ x 1+ x 2 - lim f Hx L lim f Hx L f Hx L 4 -2 HDNEL 2 x -3 x -1 x 1 x 2 + lim f Hx L lim f Hx L DNE left right HDNEL 2 lim f Hx L lim f Hx L 2 x 2 lim f Hx L (1 continued) b) Using the definition of asymptotes, explain why x = 1 is a vertical asymptote, but x = -1 is not. Vert. Asymps( lim x a f Hx L = ): x = 1 satisfies the definition; f Hx L is finite near x = -1 (1 continued) b) Using the definition of asymptotes, explain why x = 1 is a vertical asymptote, but x = -1 is not. Vert. Asymps( lim x a c) Give the x -values for which f is not continuous. Explain the kind of discontinuity of each. At x = -1 left and right limits exist but are unequal(jump discontinuity); at x = 1 the limit does not ity). f is continuous at x = -3 exist; at x = 2 left and right limits exist and are equal but are not equal to f H2L (removable discontinu2. [8] Can f Hx L approach a limit as x c even if f Hc L is undefined? If so, give an example. sinHx L Yes, f Hx L = with c=0 is an example. lim f Hx L = 1, but f is undefined at x = 0. f Hx L = ): x = 1 satisfies the definition; f Hx L is finite near x = -1 x x 0 logHx +1L x 3. [8] If we know that 1 - 1 + x for x .5, then show how to use the Squeeze Theo rem to find lim . lim f Hx L = b . x a The Squeeze Theorem tells us that if g1 Hx L f Hx L g2 Hx L near a and lim g1 Hx L = b = lim g2 Hx L, then x a x a logHx +1L x x 0 2 x Let x lim g1 Hx L = lim I1 - M = 1 = lim H1 + x L = lim g2 Hx L, then lim f Hx L = 1. 2 x 0 x g1 Hx L = 1 - x 0 and 2 x 0 g2 Hx L = 1 + x , x1 x>1 a=0, we see that since x 0 x 0 4. [8] Let f Hx L 2x 2 x +2ax +2 . What should a be to make the function continuous? x 1+ f Hx L will be continuous at x = 1 if lim f Hx L = lim Ix 2 + 2 a x + 2M = 3 + 2 a and lim f Hx L = lim 2 x = 2, exist and are equal. We need 3 + 2 a = 2, or a x 1+ 1 = - . 2 x 1x 1- Determine lim f Hx L and lim f Hx L. x 1+ x 1- 5. [8] If lim f H g Hx LL exists, must lim g Hx L also exist? Explain or give a counter-example. x 0 x 0 x 1 1 1 all what g Hx L is! Consider also f Hx L = g Hx L = ; f H g Hx LL = = = x , which gives another counter-ex1 ample. g 's limit need not exist. Suppose, for example, f Hx L = 1; the first limit exists and it doesn't matter at g Hx L x 6. [28] Evaluate the following limits using +, - and "does not exist" where applicable. Use only algebraic methods (i.e., no tables). Justify your answers. H2 x +3L 2 x2 2 x +3 7 a) lim x -6 = lim Hx -2L = lim = 2 x 2 3 x -7 x +2 b) Identify the horizontal and vertical asymptotes of the function in a) 1 Using the factorization and result of a), we see that x = 2 is not an asymptote, but x = is since the -55 1 2 x 2 -x -6 2 2 3 x 2 H3 x -1L Hx -2L x 2 3 x -1 5 6. [28] Evaluate the following limits using +, - and "does not exist" where applicable. Use only algebraic methods (i.e., no tables). Justify your answers. H2 x +3L 2 x2 2 x +3 7 a) lim x -6 = lim Hx -2L = lim = 2 x 2 3 x -7 x +2 b) Identify the horizontal and vertical asymptotes of the function in a) 1 Using the factorization and result of a), we see that x = 2 is not an asymptote, but x = is since the x 2 H3 x -1L Hx -2L 1 . 3 x 2 3 x -1 5 numerator is c) lim 2 x 2 - d) x 1 x -1 x +1 lim = x 1+ x -1 h0 h x +1 lim Hh+1L3 -1 h3 +3 h2 +3 h+1 - 1 h3 +3 h2 e) lim = lim = lim +3 h = lim Ih2 + 3 h + 3M = 3 (because x - 2 = -Hx - 2L when x < 2) 3 x +4H -2L = lim - x = lim - 3 x + 4 = -10. x -2 x 2 x 2 Check right and left limits: h0 = 3 x -2 x -8 x -2 -55 9 0 2 x2 2 2 at x = Since lim x -6 = , y = is the only 3 x 2 -7 x +2 3 3 x ر H3 x +4L H x -2L 3 x +4H -2L 3 x +4 x -2 lim = lim = lim - x x -2 x -2 x -2 x 2 x 2 x 2 horizontal asymptote. 3 x +1 lim x 1- x -1 = - since the denominator is negative and since the denominator is positive. The limit does not exist. f) g) Let f Hx L = x a of a and the limit. x -2 lim !!!!!!!!!! x -1 -1 x 2 x +a x lim f Hx L = lim = lim +a . We can see that the limit of the denominator will be 2 zero () and therefore the whole limit will not exist unless possibly the limit of the numerator is also zero. The limit of the numerator: x a 2 x a x - Ha - 1L x - a h h h0 h0 !!!!!!!!!! !!!!!!!!!! !!!!!!!!!!! Hx -2L x -1 +1 x -2 x -1 +1 = lim = lim !!!!!!!!!! = lim x - 1 + 1 = 2 !!!!!!!!!! Hx -1L-1 x -1 -1 x -1 +1 x 2 x 2 x 2 x 2 +a . Are there any values of a such that lim f Hx L exists? 2 x - Ha - 1L x - a x a 2 If so, give the value x a Hx - a L Hx + 1L 2 x x If a = 0 lim f Hx L = lim = lim = 0 2 If a = -1 lim f Hx L = x 0 x -1 lim Ix 2 + a M = a 2 + a = a Ha + 1L is only zero if a = 0 or -1. x 0 x +x x 0 x +1 Hx -1L Hx +1L x 2 -1 lim = lim 2 2 Hx +1L x -1 x +2 x +1 x -1 x -1 = lim Does not exist (see d)). x -1 x +1 Summing up, if a = 0, the limit is 0, otherwise the limit does not exist. 7. [7] Show that f Hx L 2 x 3 + x 2 + 2 has a root in H -2, -1L. Name any theorem(s) used and show all steps necessary. (Do not attempt to find a decimal approximation to the root.) We use the Intermediate Value Theorem. The function is continuous; f H-2L = -10 and f H-1L = 1. Zero is an intermediate value between -10 and 1, so by the theorem there exists an x* oe H-2, -1L such that f Hx* L = 2 x* 3 + x* 2 + 2 = 0 as requested. choice. Since: 5 x - 10 < 0.2 if and only if 5 x - 2 < 0.2 if and only if x - 2 < .04 8. [8] What is the largest value of d that guarantees " x - 2 < d implies 5 x - 10 < 0.2 "? Justify your Hence d = .04 guarantees that x - 2 < d implies 5 x - 10 < 0.2 and no smaller d will do. 8. [8] What is the largest value of d that guarantees " x - 2 < d implies 5 x - 10 < 0.2 "? Justify your choice. Since: 5 x - 10 < 0.2 if and only if 5 x - 2 < 0.2 if and only if x - 2 < .04 9. [9] Let f Hx L The f H4+hL- H4L lim f 4+h-4 h0 Hence d = .04 guarantees that x - 2 < d implies 5 x - 10 < 0.2 and no smaller d will do. a) Find the slope of the secant line through the points H4, f H4LL and H5, f H5LL. !!!! !!!! f H5L-f 5 -2 The slope of the secant is H4L = = 5 - 2 5-4 1 b) Find the slope of the tangent line at the point H4, f H4LL by evaluating a limit (show work). slope of c) Find the equation of the tangent line through H4, f H4LL. 4 1 1 Using the point-slope formula at (4,2): y - 2 = Hx - 4L fl y = x + 1 4 !!!!!!!!!! 4+h = lim -2 h h0 !!!! x. the tangent = !!!!!!!!!! 4+h lim -2 h h0 line at !!!!!!!!!! 4+h +2 !!!!!!!!!! 4+h +2 (4,f(4)) H4+hL-4 = lim !!!!!!!!!! h0 h 4+h +2 is by 1 1 = lim = !!!!!!!!!! definition h0 4+h +2 4 ...
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