Math 105 Practice Exam #1 Fall 2013 Answer Key

Math 105 1. Practice Exam #1 Fall 2013 ANSWER KEY Consider the line L given by x + 3y = 6. (a) Sketch this line L. 1 x + 3y = 6 =⇒ 3y = −x + 6 =⇒ y = − x + 2 3 See graph below. (b) Find the equation of the new line M that is perpendicular to the first line L, x + 3y = 6, and passes through the point (1, −2). 1 Line L has slope equal to − . Then the perpendicular slope is opposite reciprocal equaling 3. 3 Then the new line M with slope 3 passing through the point (1, −2) is given by y − (−2) = 3(x − 1) or y = 3x − 5 (c) Sketch this new line M found in (b). See graph below. Both (a) and(c) are on the same graph here. 2. Evaluate each of the following limits. Please justify your answers. Be clear if the limit equals a value, +∞ or −∞, or Does Not Exist. 1 (a) x2 + 5x − 14 DSP = x2 − 4x + 4 lim x→−7 0 = 0 81 x2 − 9x + 20 = DOES NOT EXIST , RHL6=LHL |4 − x| (b) lim x→4 RHL: lim x→4+ x2 − 9x + 20 (x − 4)(x − 5) x2 − 9x + 20 00 DSP = lim = lim = lim x − 5 = −1 + + + |4 − x| −(4 − x) x − 4 x→4 x→4 x→4 (x − 4)(x − 5) x2 − 9x + 20 x2 − 9x + 20 00 DSP = lim = lim −(x − 5) = 1 = lim − − − |4 − x| 4−x −(x − 4) x→4 x→4 x→4 LHL: lim x→4− Here, recall that |4 − x| =   4−x  −(4 − x) if 4−x≥0 if 4−x<0 = Warning: watch the order here on 4 − x. It’s not x − 4. (c) f (x2 ) + 5x − 8 = [f (x)]2 + 5x + 14 lim x→−6   4−x  x−4 if x ≤ 4 ← LHL case if x > 4 ← RHL case where f (x) = x + 2 0 x2 + 5x − 6 f (x2 ) + 5x − 8 0 x2 + 2 + 5x − 8 = lim = lim x→−6 x2 + 4x + 4 + 5x + 14 x→−6 [x + 2]2 + 5x + 14 [f (x)]2 + 5x + 14 lim x→−6 7 −7 (x + 6)(x − 1) x−1 x2 + 5x − 6 = lim = lim = = x2 + 9x + 18 x→−6 (x + 6)(x + 3) x→−6 x + 3 −3 3 = lim x→−6 0 (x + 7)(x − 2) x2 + 5x − 14 0 x+7 = lim DOES NOT EXIST , RHL6=LHL = lim x→2 (x − 2)(x − 2) x→2 x − 2 x2 − 4x + 4 (d) lim x→2 RHL: lim x→2+ LHL: lim x→2− (e) lim x→8 = lim x→8 = lim x→8 x+7 9 = + = +∞ x−2 0 x+7 9 = − = −∞ x−2 0 0 √ √ √   3− x+1 0 3− x+1 3+ x+1 9 − (x + 1) √ √ · = lim = lim 2 x→8 x2 − 7x − 8 x→8 x2 − 7x − 8 3+ x+1 (x − 7x − 8)(3 + x + 1) −(x − 8) 8−x √ √ = lim (x − 8)(x + 1)(3 + x + 1) x→8 (x − 8)(x + 1)(3 + x + 1) −1 √ = (x + 1)(3 + x + 1) (f ) lim x→4 3−x 3 − x−5 7−x x2 − x − 12 0 0 = lim x→4 −1 √ = (9)(3 + 9)  −1 = (9)(6) −1 54 (3 − x)(7 − x) − 3(x − 5) (x − 5)(7 − x) x2 − x − 12 2     2 21 − 10x + x2 − 3x + 15 x − 13x + 36 (x − 5)(7 − x) (x − 5)(7 − x) = lim x→4 x2 − x − 12 x2 − x − 12     x2 − 13x + 36 1 1 (x − 9)(x − 4) · · = lim x→4 (x − 5)(7 − x) (x − 5)(7 − x) x2 − x − 12 (x − 4)(x + 3)  = lim x→4 = lim x→4 = lim x→4 5 4−9 −5 x−9 = = = (x − 5)(7 − x)(x + 3) (4 − 5)(7 − 4)(4 + 3) (−1)(3)(7) 21 x−5 DNE b/c RHL 6= LHL = x+3 x−5 −8 RHL: lim = + = −∞ + 0 x→−3 x + 3 −8 x−5 = − = +∞ LHL: lim − 0 x→−3 x + 3 (g) lim x→−3 (h) lim − x→8 1 = −∞ b/c RHL=LHL (x − 8)2 RHL: lim − x→8+ LHL: lim − x→8− 1 −1 = + = −∞ 2 (x − 8) 0 −1 −1 1 = − 2 = + = −∞ 2 (x − 8) (0 ) 0 x+7 f (x + h) − f (x) . Compute the difference quotient . Simplify x−3 h your answer until the h in the denominator cancels.   (x + h + 7)(x − 3) − (x + 7)(x + h − 3) (x + h) + 7 x + 7 − f (x + h) − f (x) (x + h − 3)(x − 3) (x + h) − 3 x − 3 = = h h h 2 2 x + xh + 7x − 3x − 3h − 21 − (x + xh − 3x + 7x + 7h − 21) = h(x + h − 3)(x − 3) 3. Suppose that f (x) = = x2 + xh + 7x − 3x − 3h − 21 − x2 − xh + 3x − 7x − 7h + 21 h(x + h − 3)(x − 3) = −10 −10h −3h − 7h = = h(x + h − 3)(x − 3) h(x + h − 3)(x − 3) (x + h − 3)(x − 3) 1+x 1 and g(x) = . 1−x x (a) Compute f ◦ g(x). Simplify your answer to a single fraction. State the Domain. 4. Consider the two functions f (x) = x 1 x+1     1+ 1 + x x+1 x+1 1 x x x = · = = = x = f ◦ g(x) = f (g(x)) = f 1 x 1 x−1 x x x−1 x−1 − 1− x x x x Domain f ◦ g: {x|x 6= 0, 1}. Recall, for x to be in the domain of f ◦ g, it must be in the domain of 3 g first, and THEN the output g(x) must be in the domain of f . (b) Compute g ◦ f (x). Simplify your answer to a single fraction. State the Domain.   1−x 1 1+x = = g ◦ f (x) = g(f (x)) = g 1 + x 1−x 1+x 1−x Domain g ◦ f : {x|x 6= −1, 1} (c) Compute f ◦ f (x). Simplify your answer to a single fraction. State the Domain. f ◦ f (x) = f (f (x)) = f  1+x 1−x  1−x 1+x 1−x+1+x 1+x + 1−x = 1−x 1−x = 1−x = 1+x 1−x 1+x 1 − x − (1 + x) 1− − 1−x 1−x 1−x 1−x 1+ 1−x+1+x 2   2 1 1−x 1 − x 1 − x = = = · = − −2x 1 − x − 1 − x) 1−x −2x x 1−x 1−x Domain f ◦ f : {x|x 6= 0, 1} 5. Consider the function defined by f (x) =  2       √   x−7          1   if x ≥ 11 if 7 < x < 11 if x = 7  7−x         16 − x2           1 x+4 if 0 < x < 7 if − 4 < x ≤ 0 if x < −4 (a) Carefully sketch the graph of f (x). 4 (b) State the Domain of the function f (x). Domain= {x : x 6= −4} (c) Compute lim f (x) = x→−4 RHL: LHL: DOES NOT EXIST since RHL 6= LHL lim f (x) = lim 16 − x2 = 0 x→−4+ x→−4+ lim f (x) = lim x→−4− x→−4− (d) Compute lim f (x) = x→0 1 1 = − = −∞ x+4 0 DOES NOT EXIST since RHL 6= LHL RHL: lim f (x) = lim 7 − x = 7 x→0+ x→0+ LHL: lim f (x) = lim 16 − x2 = 16 x→0− x→0− (e) Compute lim f (x) = 0 since RHL=LHL x→7 √ RHL: lim f (x) = lim x − 7 = 0 x→7+ x→7+ 5 LHL: lim f (x) = lim 7 − x = 0 x→7− x→7− (f ) Compute lim f (x) = 2 x→11 RHL: lim f (x) = lim 2 = 2 x→11+ x→11+ LHL: lim f (x) = lim x→11− x→11− √ x−7= √ 4=2 (g) State the value(s) at which f is discontinuous. Justify your answer(s) using the definition of continuity discussed in class. • f is discontinuous at x = 7, because despite the fact that f (7) = 1 is defined, and lim f (x) = 0, x→7 those two values are not equal. • f is discontinuous at x = 0, because despite the fact that f (0) = 16 is defined, the lim f (x) x→0 DOES NOT EXIST. • f is discontinuous at x = −4 for two reasons, f (−4) is undefined, and the lim f (x) DOES NOT x→−4 EXIST. Note that f is continuous at x = 11 because lim f (x) = 2 = f (11) x→11 6