Limits and Derivatives Calculation Exercise - Prof. Gary D. Berg, Quizzes of Mathematics

Download Limits and Derivatives Calculation Exercise - Prof. Gary D. Berg and more Quizzes Mathematics in PDF only on Docsity! haas (ah28474) – HW01 – BERG – (56495) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine if lim x→ 7 √ x + 9 − 4 x − 7 exists, and if it does, find its value. 1. limit = 8 2. limit = 7 3. limit does not exist 4. limit = 1 4 5. limit = 1 8 correct 6. limit = 1 9 Explanation: Since ( √ x + 9 − 4)( √ x + 9 + 4) = (x + 9) − 16 = x − 7 , we see by rationalizing the numerator that √ x + 9 − 4 x − 7 = x − 7 (x − 7)( √ x + 9 + 4) = 1√ x + 9 + 4 provided x 6= 7. On the other hand, lim x→ 7 √ x + 9 + 4 = 8. Consequently, by properties of limits, lim x→ 7 √ x + 9 − 4 x − 7 exists and has limit = 1 8 . 002 10.0 points Find the value of lim x→−4 3 x + 4 ( 4 x2 + 8 − 1 6 ) . 1. limit = 1 12 2. limit = 1 4 3. limit does not exist 4. limit = 1 6 correct 5. limit = 1 8 Explanation: After the second term in the product is brought to a common denominator it becomes 24 − x2 − 8 6(x2 + 8) = 16 − x2 6(x2 + 8) . Thus the given expression can be written as 3(16 − x2) 6(x + 4)(x2 + 8) = 3(4 − x) 6(x2 + 8) so long as x 6= −4. Consequently, lim x→−4 3 x + 4 ( 4 x2 + 8 − 1 6 ) = lim x→−4 3(4 − x) 6(x2 + 8) . By properties of limits, therefore, limit = 1 6 . 003 10.0 points haas (ah28474) – HW01 – BERG – (56495) 2 Find the derivative of f when f(x) = 2 + cos x sin x . 1. f ′(x) = −1 + 2 cos x sin2 x correct 2. f ′(x) = 2 + sin x cos2 x 3. f ′(x) = −2 + cos x sin2 x 4. f ′(x) = 2 sin x + 1 cos2 x 5. f ′(x) = 2 sin x − 1 cos2 x 6. f ′(x) = sin x − 2 cos2 x 7. f ′(x) = 1 − 2 cosx sin2 x 8. f ′(x) = 2 − cos x sin2 x Explanation: By the quotient rule, f ′(x) = − sin2 x − cos x(2 + cos x) sin2 x = − sin2 x − cos2 x − 2 cos x sin2 x . But cos2 x + sin2 x = 1. Consequently, f ′(x) = −1 + 2 cosx sin2 x . 004 10.0 points Find the derivative of f when f(x) = 3x cos 2x − 8 sin 2x . 1. f ′(x) = −6x sin 2x − 16 cos 2x 2. f ′(x) = −16 cos 2x + 6x sin 2x 3. f ′(x) = 6x sin 2x − 13 cos 2x 4. f ′(x) = −6x sin 2x−13 cos 2x correct 5. f ′(x) = 16 cos 2x − 13x sin 2x Explanation: Using formulas for the derivatives of sine and cosine together with the Product and Chain Rules, we see that f ′(x) = 3 cos 2x − 6x sin 2x − 16 cos 2x = −6x sin 2x − 13 cos 2x . 005 10.0 points Find f ′(x) when f(x) = 1√ 6x − x2 . 1. f ′(x) = 3 − x (x2 − 6x)3/2 2. f ′(x) = x − 3 (6x − x2)3/2 correct 3. f ′(x) = 3 − x (x2 − 6x)1/2 4. f ′(x) = x − 3 (6x − x2)1/2 5. f ′(x) = 3 − x (6x − x2)3/2 6. f ′(x) = x − 3 (x2 − 6x)3/2 Explanation: By the Chain Rule, f ′(x) = − 1 2(6x − x2)3/2 (6 − 2x) . Consequently, f ′(x) = x − 3 (6x − x2)3/2 . haas (ah28474) – HW01 – BERG – (56495) 5 it thus follows by properties of limits that lim x→∞ √ 1 + 8 x + √ 1 − 6 x exists and has value 2. Consequently, again by properties of limits, the limit lim x→∞ √ x ( √ x + 8 − √ x − 6) exists and limit = 7 . 011 10.0 points Determine if the limit lim x→∞ ( 1 3 − e x 4ex + 3 ) exists, and if it does, compute its value. 1. limit = − 1 12 2. limit does not exist 3. limit = 1 3 4. limit = 1 12 correct 5. limit = −1 4 Explanation: Adding, we see that 1 3 − e x 4ex + 3 = ex + 3 3 (4ex + 3) , and so 1 3 − e x 4ex + 3 = 1 + 3e−x 3 (4 + 3e−x) after dividing by ex in both numerator and denominator. But lim x→∞ e−x = 0 , in which case lim x→∞ ( 1 + 3e−x ) = 1 , while lim x→∞ 3 ( 4 + 3e−x ) = 12 . Consequently, by properties of limits, the given limit exists and limit = 1 12 . 012 10.0 points Find the derivative of f when f(x) = e2x+5 + 3e−2x+5. 1. f ′(x) = 2e5 (e2x − 3e−2x) correct 2. f ′(x) = 2(e2x − 3e−2x) 3. f ′(x) = e5 (e2x + 3e−2x) 4. f ′(x) = e5 (e2x − 3e−2x) 5. f ′(x) = 2(e2x + 3e−2x) 6. f ′(x) = e5 (3e2x − e−2x) Explanation: After differentiation, f ′(x) = 2e2x+5 − 6e−2x+5. Consequently, f ′(x) = 2e5 (e2x − 3e−2x) . 013 10.0 points Determine f ′(x) when f(x) = e √ 5x+6. 1. f ′(x) = 5 2 e √ 5x+6 √ 5x + 6 correct haas (ah28474) – HW01 – BERG – (56495) 6 2. f ′(x) = 5 2 e √ 5x+6 √ 5x + 6 3. f ′(x) = 1 2 e √ 5x+6 √ 5x + 6 4. f ′(x) = 5e √ 5x+6 5. f ′(x) = 5e √ 5x+6 √ 5x + 6 Explanation: By the chain rule f ′(x) = e √ 5x+6 ( d dx √ 5x + 6 ) = 5 2 e √ 5x+6 √ 5x + 6 . 014 10.0 points Find the derivative of f when f (θ) = ln (cos 3θ) . 1. f ′ (θ) = − 1 sin 3θ 2. f ′ (θ) = −3 cot 3θ 3. f ′ (θ) = 3 cos 3θ 4. f ′ (θ) = cot 3θ 5. f ′ (θ) = −3 tan 3θ correct 6. f ′ (θ) = 3 tan 3θ Explanation: By the Chain Rule, f ′(θ) = 1 cos(3θ) d dθ (cos 3θ) = −3 sin 3θ cos 3θ . Consequently, f ′(θ) = −3 tan 3θ . 015 10.0 points Differentiate the function f(x) = cos(ln 2x) . 1. f ′(x) = −2 sin(ln 2x) x 2. f ′(x) = sin(ln 2 x) x 3. f ′(x) = 1 cos(ln 2 x) 4. f ′(x) = −sin(ln 2 x) x correct 5. f ′(x) = 2 sin(ln 2x) x 6. f ′(x) = − sin(ln 2 x) Explanation: By the Chain Rule f ′(x) = −sin(ln 2x) x . 016 10.0 points Determine f ′(x) when f(x) = e(4 ln(x 2)) . 1. f ′(x) = 1 x e4 ln(x 2) 2. f ′(x) = e8/x 3. f ′(x) = 7x8 4. f ′(x) = 8x7 correct 5. f ′(x) = 8(lnx)e4 ln(x 2) haas (ah28474) – HW01 – BERG – (56495) 7 6. f ′(x) = 4 x2 e4 ln(x 2) Explanation: Since r lnx = lnxr , eln x = x , we see that f(x) = e(ln x 8) = x8 . Consequently, f ′(x) = 8x7 . 017 10.0 points Find the derivative of f when f(x) = 4 (2x) − 8 log2 x . 1. f ′(x) = 4 (2x) ln 2 − 8 ln 2 x 2. f ′(x) = 4 (2x) ln 2 − 8 log2 x 3. f ′(x) = 4 (2x) ln 2 − 8 x 4. f ′(x) = 4 (2x) − 8 x ln 2 5. f ′(x) = 4 (2x) ln 2 − 8 x ln 2 correct Explanation: Note that 2x = ex ln 2, log2 x = lnx ln 2 . By the chain rule, therefore, f ′(x) = 4ex ln 2 ln 2 − 8 x ln 2 . Consequently, f ′(x) = 4 (2x) ln 2 − 8 x ln 2 . 018 10.0 points Find the nth-derivative of f when f(x) = 3 ln(4x + 3). 1. f (n)(x) = (−1)n−1(n − 1)! 3 · 4 n (4x + 3)n correct 2. f (n)(x) = (−1)n−1(n − 1)! 3 (4x + 3)n 3. f (n)(x) = (n − 1)! 3 · 4 n (4x + 3)n 4. f (n)(x) = (−1)n(n − 1)! 3 · 4 n (4x + 3)n 5. f (n)(x) = (−1)n−1n! 3 · 4 n (4x + 3)n Explanation: Applying the Chain Rule successively we see that f (1)(x) = 12 4x + 3 , f (2)(x) = − 48 (4x + 3)2 , and f (3)(x) = 384 (4x + 3)3 , f (4)(x) = −2·3 3 · 4 4 (4x + 3)4 Continuing the successive application of the Chain Rule we see that f (4)(x) = −2 · 3 3 4 4 (4x + 3)4 = − 3! 3 · 4 4 (4x + 3)4 , f (5)(x) = 2 · 3 · 4 3 4 5 (4x + 3)5 = 4! 3 · 45 (4x + 3)5 , and so on. Thus f (n)(x) =        −(n − 1)! 3 · 4 n (4x + 3)n , n even, (n − 1)! 3 · 4 n (4x + 3)n , n odd.