GTK Mathematics II Midterm 1 PDF

GTK Mathematics II Midterm 1 extra material Miskei-Váradi Ferenc István 2024 1. Determine the derivative of the following functions. Give the equation
of the tangent line to the graph at the point a, f (a) (linear approximation around x = a). a) b) c) d) e) f) g) h) i) j) f1 (x) = 5x 7 − 12x, a = 1...

GTK Mathematics II Midterm 1 extra material Miskei-Váradi Ferenc István 2024 1. Determine the derivative of the following functions. Give the equation
of the tangent line to the graph at the point a, f (a) (linear approximation around x = a). a) b) c) d) e) f) g) h) i) j) f1 (x) = 5x 7 − 12x, a = 1 f2 (x) = x23 , a = −2 √ f3 (x) = 2 x a = 14 f4 (x) = (3x − 5)(8 − 2x), a = 2 f5 (x) = −5x−7 2x+8 , a = −3 f6 (x) = 42 − 76x , a = 0 f7 (x) = (2 − 3x)4 , a = 2 2 f8 (x) = e x −2x−15 , a = −3 f9 (x) = e 2x ln (3x − 5), a = 2 f10 (x) = (5x 2 + 6x − 7)(8x 3 − 9x 2 + 100), a = −2 2. Determine the monotonicity and extreme values of the following functions. a) b) c) d) e) f) g) h) f1 (x) = −8x f2 (x) = x 2 − 6x + 42 f3 (x) = x 3 − 243x f4 (x) = x 3 + 3x 2 + 3x + 100 f5 (x) = −3x 4 − 32x 3 − 90x 2 f6 (x) = e −x (3 − 3x − x 2 ) 1 f7 (x) = x−8 f8 (x) = x 22−1 3. Determine the convexivity and the inflection points of the following functions. a) b) c) d) e) f) g) h) f1 (x) = 21 x + 7 f2 (x) = x 2 − 6x + 42 f3 (x) = x 3 − 21x 2 f4 (x) = x 3 + 3x 2 + 3x + 100 f5 (x) = −3x 4 − 36x 3 − 126x 2 f6 (x) = e x (5x − 18) 1 f7 (x) = x−8 f8 (x) = x 22−1 4. Given the cost function C (q) and the selling price function p(q), determine the value of q for which the profit P(q) = R(q) − C (q). a) C (q) = 0.02q 2 + 4q, p(q) = 6 + 0.01q b) C (q) = 0.01q 2 + q + 2, p(q) = 42 − 0.015q c) C (q) = 0.05q 2 + 5q + 10, p(q) = 12.5 − 0.075q 5. Determine the elasticity of the following functions. a) b) c) d) f1 (x) = e 5x f2 (x) = x 2 f3 (x) = 0.01x 2 + 5x + 10 f4 (x) = ln x 6. The demand for a given product at unit price of p dollars is D(p) = 100 − 0.005p 2 units. a) How much revenue can we generate if we set a 20 dollar price? b) What’s the elasticity of the demand at the 20 dollar/unit price? c) By what percentage will the raising of the price by 2% change the demand (starting from p = 20) (Use the value of elasticity.) d) At what price would we have maximal profit? 7. Given the cost function C (q) and the selling price function p(q), determine the revenue and the marginal profit at the price q. a) C (q) = 0.02q 2 + 4q, p(q) = 42 − 0.015q, q = 10 b) C (q) = 0.05q 2 + 5q + 10, p(q) = 12.5 − 0.075q, q = 20 c) C (q) = 0.1q 2 + 6q + 100, p(q) = 6 + 0.04q, q = 25 Solutions 1. f1′ (x) = 35x 6 − 12, f1′ (1) = 23, L1 (x) = 23(x − 1) − 7 3 −3 1 ′ f2′ (x) = −6 x 4 , f2 (−2) = − 8 , L2 (x) = 8 (x +
2) − 4 1 1 ′ ′ f3 (x) = √x , f3 (1/4) = 2, L3 (x) = 2 x − 4 + 1 f4′ (x) = −12x + 34, f4′ (2) = 10, L4 (x) = 10(x − 2) + 4 −13 −13 f5′ (x) = 2x 2 +16x+32 , f5′ (−3) = −13 2 , L5 (x) = 2 (x + 3) + 4 ′ 6x ′ f6 (x) = −6 ln 7 · 7 , f6 (0) = −6 ln 7, L6 (x) = −6 ln 7x + 41 f7′ (x) = −12(2 − 3x)3 , f7′ (2) = 768, L7 (x) = 768(x − 2) + 256 2 f8′ (x) = (2x − 2)e x −2x−15 , f8′ (−3) = −8, L8 (x) = −8(x + 3) + 1 3 i) f9′ (x) = 2e 2x ln (3x − 5) + e 2x 3x−5 , f9′ (2) ≈ 163.79, L9 (x) = 163.79(x − 2) ′ j) f10 (x) = (10x+6x)(8x 3 −9x 2 +100)+(5x 2 +6x−7)(24x 2 −18x), ′ f10 (−2) = 132, L10 (x) = 132(x + 2) a) b) c) d) e) f) g) h) 2. a) f1 (x) = −8x −→ f1′ (x) = −8 < 0 everywhere, −→ f1 strictly decr. everywhere. f1 has no extreme values b) f2 (x) = x 2 − 6x + 42 −→ f2′ (x) = 2x − 6 −→ f2 strictly decr. on (−∞, 3], strictly incr. on [3, +∞). At x = 3, it has a strict global minimum, f2 (3) = 33. c) f3 (x) = x 3 − 243x −→ f3′ (x) = 3x 2 − 243 = 3(x + 9)(x − 9) −→ strictly incr. on (−∞, −9], strictly decr. on [−9, +9], strictly incr. on [9, +∞). x = −9 is a strict local maximum location, f3 (−9) = 1458, x = 9 strict local minimum location f3 (9) = −1458. d) f4 (x) = x 3 + 3x 2 + 3x + 100 −→ f4′ (x) = 3x 2 + 6x + 3 = 3(x + 1)2 ⩾ 0. f4 strictly incr. on R. e) f5 (x) = −3x 4 − 32x 3 − 90x 2 −→ f5′ (x) = −12x 3 − 96x 2 − 180x = −12x(x 2 + 8x + 15) = −12x(x + 5)(x + 3) −→ f5 strictly incr. on (−∞, −5], strictly decr. on [−5, −3], strictly incr. on [−3, 0], and strictly decr. on [0, +∞). x = −5 is a strict local maximum location f5 (−5) = −125, x = −3 strict local minimum location f5 (−3) = −189, x = 0 strict global max, f5 (0) = 0. 2. f) f6 (x) = e −x (3 − 3x − x 2 ) −→ f6′ (x) = e −x (x 2 + x − 6) = 0 ⇐⇒ 0 = x 2 + x − 6 ⇐⇒ x = 2 or x = −3. x = −3 is a strict global maximum location, f (−3) = 3e 3 , x = 2 is a strict local minimum f (2) = −7/e 2. −1 1 −→ f7′ (x) = (x−8) g) f7 (x) = x−8 2 < 0 =⇒ f7 strictly decreases on (−∞, 8) and f7 strictly decr. on (8, +∞). f7 has no extreme values. h) f8 (x) = x 22−1 −→ f8′ (x) = (x−4x 2 −1)2 is positive on (−∞, −1) and on (−1, 0). In 0 it changes signs (+) → (−). f8′′ is negative on (0, 1) and on (1, +∞). Hence f8 is strictly incr. on (−∞, 1), strictly incr. on (−1, 0], strictly decr. on [0, 1), and strictly decr. on (1, +∞). f8 has a strict local maximum at 0. 3. a) f1 (x) = 21 x + 7 −→ f1′′ (x) = 0 is everywhere non-negative and everywhere non-positive, so f1 is everywhere (weakly) concave up and (weakly) concave down. b) f2 (x) = x 2 − 6x + 42 −→ f2′′ (x) = 2 > 0 everywhere. so f2 everywhere (strictly) concave up. c) f3 (x) = x 3 − 21x 2 −→ f3′′ (x) = 6x − 42 = 0 ⇐⇒ x = 7. On (−∞, 7) f ′′ < 0, so f3 is strictly concave down on (−∞, 7], and on (7, +∞) f ′′ > 0, so f3 is strictly concave up on [7, +∞). x = 7 is an inflection point. d) f4 (x) = x 3 + 3x 2 + 3x + 100 −→ f4′′ (x) = 6x + 6. On (−∞, −1) f ′′ < 0, so f3 is strictly concave down on (−∞, −1]. On (−1, +∞), we have f ′′ > 0, so f3 is strictly concave up on [−1, +∞). x = 1 is an inflection point e) f5 (x) = −3x 4 − 36x 3 − 126x 2 −→ √ 2 f5′′ (x) = −36x − 252 = 0 √ ⇐⇒ x = √ −3 ± 2. On √ − 216x ′′ (−∞, −3 − 2) f5′′ > 0 and √ 2) f5 0. Therefore f6 is strictly concave down on (−∞, 8/5], strictly concave up on [8/5, +∞) and x = 8/5 is an inflection point. 2 1 ′′ −→ f7′′ (x) = (x−8) g) f7 (x) = x−8 3. On (−∞, 8) f7 < 0, ′′ (8, +∞)-n f7 > 0. So on (−∞, 8) f7 is strictly concave down, (8, +∞)-n f7 is strictly concave up. x = 8 is not inflection point, because the function is not defined there. 2 +1) h) f8 (x) = x 22−1 −→ f8′′ (x) = 4(3x (x 2 −1)3 (after simplification). The enumerator is always positive, so we need only examine the sign of the denominator. Based on which, on (−∞, 1) f8′′ > 0, on (−1, 1) f8′′ < 0, and on (1, +∞) f8′′ > 0. So on (−∞, 1) f8 is strictly concave up, on (−1, 1) f8 is strictly concave down, and on (1, +∞) f8 is strictly concave up. 4. a) P(q) = qp(q) −C (q) = q(6 + 0.01q) − (0.02q 2 + 4q) = | {z } R(q) = −0.01q 2 + 2q =⇒ P ′ (q) = −0.02q + 2 = 0 ⇐⇒ x = 10 Therefore the profit is maximal at the sale of q = 100 products, P(100) = 100. b) R(q) = qp(q) = q(42−0.015q) = 42q −0.015q 2.Maximization / minimization of the profit is equivalent to solving the equation C ′ (q) = R ′ (q).. C ′ (q) = 0.02q + 1, R ′ (q) = 42 − 0.03q. 0.02q + 1 = 42 − 0.03q ⇐⇒ q = 820. Therefore the profit is maximal at the sale of q = 820 products, P(820) = 16808. 4. c) P ′ (q) = ( qp(q) −C (q))′ = (q(12.5 − 0.075q))′ − (0.1q + 5) = | {z } R(q) = 7.5 − 0.25q = 0 ⇐⇒ q = 30. Therefore the profit is maximal at the sale of q = 30 products, P(30) = 102.5. 5. The formula for the elasticity is E (x) = xf ′ (x) f (x) a) f1 (x) = e 5x =⇒ E (x) = x(5e 5x ) = 5x e 5x b) f2 (x) = x 2 =⇒ E (x) = x(2x) =2 x2 c) f3 (x) = 0.01x 2 + 5x + 10 =⇒ E (x) = x(0.02x + 5) 0.01x 2 + 5x + 10 d) f4 (x) = ln x =⇒ E (x) =
x x1 1 = ln x ln x 6. D(p) = 100 − 0.005p 2 a) At a unit price of 20 dollars, we can sell D(20) = 98 units of product. ′ (p) −0.01p 2 b) E (p) = pD D(p) = 100−0.005p 2 , E (20) ≈ −0.0408163 c) by ≈ 2 · E (20) ≈ −0.0816326 percent. (The actual change is D(2.04) − D(2) = −0.000808, which is ≈ −0.08165 percent of the value D(2) = 99.98.) d) R(p) = pD(p) = 100p − 0.005p 3. We are looking for a place where the function R ′ (p) changes signs (+) → (−). R ′ (p) = 100 − 0.015p 2 = 0 ⇐⇒ p ≈ ±81.6497. P ′ changes (+) → (−) sign at p ≈ +81.6497 =⇒ the optimal solution is p ≈ +81.6497. 7. The marginal cost is C ′ (.), the profit is P(q) = qp(q) − C (q). a) C ′ (10) =0.04 · 10 + 4 = 4.4 P(10) =10 · 41.85 − 42 = 376.5 b) C ′ (20) =0.1 · 20 + 5 = 7 P(20) =20 · 11 − 130 = 90 c) C ′ (25) =0.2 · 25 + 6 = 11 P(25) =25 · 7 − 312.5 = −137.5