Questions and Answers #3 Functions

Questions and Answers Sheet 3 Functions Question #1 Find the exact length of the curve. 𝑦 = 𝑥3+ 1 , 1 ≤ 𝑥 ≤ 2 3 4𝑥 Answer: The function 𝑓(𝑥) = 𝑓 ′ (𝑥) = 𝑥 2 − 1 4𝑥 2 𝑥3 3 + 1 4𝑥 is continuous, differentiable on [1,2] and its derivative is continuous on [1,2] so the length L of the curve is given by the formula: L= 2 1 1 + [ x2 − 1 2 ] dx 4x2 therefore, L= 2 1 = 2 1  2  2 1 1 [ x4 + + ]dx 2 16 x 4 [ x2 + 1 1 1 1 1 + [ x4 − + ]dx 2 16 x 4 x2 + 1 2 ] dx 4 x2 1 dx 4 x2 1 1 2 3 4𝑥 1 = [ 𝑥3 − ] 1 1 1 1 ( 23 − ) − ( 13 − ) 3 4⋅2 3 4⋅1 = 8 1 1 1 − − + 3 8 3 4 = 64 3 8 6 − − + 24 24 24 24 𝐿= 59 24 Question #2 Find a function f such that 𝐹 = ∇𝑓 𝐹(𝑥, 𝑦) = 𝑥 2 𝑖 + 𝑦 2 𝑗 C is the arc of the parabola 𝑦 = 2𝑥 2 from (-1,2) to (2,8) Answer: The x component of the given field is equal to 𝑥 2 . We first integrate with respect to x. It follows that: 𝑓(𝑥, 𝑦) = 𝑥3 + 𝑔(𝑦) 3 Take a partial derivative of f with respect to y to obtain that: 𝑔′ (𝑦) = 𝑦 2 = 𝑓𝑦 Next, we integrate once again but this time with respect to y to get: 𝑓(𝑥, 𝑦) = 𝑦3 𝑥 3 + +𝐶 3 3 Set 𝐶 = 0 since it doesnt matter for the fundamental theorem to be applied. We have: 𝑓(𝑥, 𝑦) = 𝑥 3 + 𝑦3 3 Result: 𝑓(𝑥, 𝑦) = 𝑥 3 +𝑦 3 3 Question #3 What is the minimum vertical distance between the parabolas 𝑦 = 𝑥 2 + 1 and 𝑦 = 𝑥 − 𝑥 2 Answer: 𝑦1 = 𝑥 1 + 1, 𝑦2 = 𝑥 − 𝑥 2 The vertical at any point x is obtained form 𝐷 = 𝑦1 − 𝑦2 = 𝑥 2 + 1 − (𝑥 − 𝑥 2 ) = 𝑥2 + 1 − 𝑥 + 𝑥2 = 2𝑥 2 − 𝑥 + 1 If you subtract the other way the distance you get is negative because of where the graphs are situated, but the problem would work out similarly. Find where 𝐷 ′ = 0 𝐷′ (𝑥) = 0 = 4𝑥 − 1 𝑥= 1 4 1 𝐷(𝑥) = 2𝑥 2 − 𝑥 + 1 is an upward opening parabola so 𝑥 = is the minimum. 4 Plug in to get the minimum distance 1 1 2 1 𝐷( ) = 2( ) − + 1 4 4 4 = 1 1 − +1 8 4 = 1−2+8 8 = 7 8 Result: 7 8 Question #4 Determine whether𝑓’(0)exist. 1   x sin if x  0 x  0 if x = 0 Answer: Recall that: 𝑓 ′ (𝑎) = lim 𝑥→𝑎 𝑓 ′ (0) 𝑓(𝑥) − 𝑓(𝑎) 𝑥−𝑎 1 𝑥 sin ( ) − 0 𝑥 = lim 𝑥→0 𝑥−0 1 𝑓 ′ (0) = lim sin ( ) 𝑥→0 𝑥 𝑓 ′ (0) = sin(∞) = DOES NOT EXIST Result: 𝑓 ′ (0) does not exist Question #5 Write the polynomial as the product of factors that are irreducible over the rationals, 𝑓(𝑥) = 𝑥 4 + 6𝑥 2 − 27 Answer: Let us factor the given polynomial function as the product of factors that are irreducible over the rationals 𝑓(𝑥) = 𝑥 4 + 6𝑥 2 − 27 = 𝑥 4 + 9𝑥 2 − 3𝑥 2 − 27 = (𝑥 4 + 9𝑥 2 ) − (3𝑥 2 + 27) Let us factor out 𝑥 2 and -3: = 𝑥 2 (𝑥 2 + 9) − 3(𝑥 2 + 9) = (𝑥 2 − 3)(𝑥 2 + 9) Result: (𝑥 2 − 3)(𝑥 2 + 9) Question #6 Determine whether the given set of functions is linearly dependent or linearly independent on the interval (−∞, ∞) 𝑓1 (𝑥) = 1 + 𝑥, 𝑓2 (𝑥) = 𝑥, 𝑓3 (𝑥) = 𝑥 2 Answer: From Theorem we know that the function 𝑓1 , 𝑓2 , and 𝑓3 are linearly independent if and only if 𝑊(𝑓1 , 𝑓2 , 𝑓3 ) ≠ 0 f1 f 2 Where W ( f1 , f 2 , f3 ) = f1 f 2 f1 f 2 f3 f3 f3 ∴ 𝑓1 (𝑥) = 1 + 𝑥𝑓2 (𝑥) = 𝑥𝑓3 (𝑥) = 𝑥 2 1 + x x x2 W ( f1 , f 2 , f 3 ) = 1 1 2x 0 0 2 Using the properties of determinants we have W ( f1 , f 2 , f3 ) = (1 + x) 1 2x 1 2x 1 1 −x + x2 0 2 0 2 0 0 𝑊(𝑓1 , 𝑓2 , 𝑓3 ) = (1 + 𝑥)(2 − 0) − 𝑥(2 − 0) + 𝑥 2 (0 − 0) 𝑊(𝑓1 , 𝑓2 , 𝑓3 ) = (1 + 𝑥)(2) − 𝑥(2) + 𝑥 2 (0) 𝑊(𝑓1 , 𝑓2 , 𝑓3 ) = 2 + 2𝑥 − 2𝑥 + 0 → 𝑊(𝑓1 , 𝑓2 , 𝑓3 ) = 2 𝑊(𝑓1 , 𝑓2 , 𝑓3 ) ≠ 0 Thus, the given function are linearly independent Result: The given functions are linearly independent As their Wroian 𝑊(𝑓1 , 𝑓2 , 𝑓3 ) = 2 ≠ 0 Question #7 g is related to one of the parent functions Use function notation to write g in terms of f. 𝑔(𝑥) = √𝑥 + 4 + 8 Answer: Rewrite g in terms of f. 𝑔(𝑥) = 𝑓(𝑥 + 4) + 8 Result: 𝑔(𝑥) = 𝑓(𝑥 + 4) + 8 Question #8 Consider the following function 𝑓(𝑥) = ln(1 + 2𝑥) , 𝑎 = 3, 𝑛 = 3,2.6 ≤ 𝑥 ≤ 3.4 Approximate f by a Taylor polynomial with degree n at the number a. Use Taylor's Inequality to estimate the accuracy of the approximation 𝑓(𝑥) ≈ 𝑇𝑛 (𝑥) when x lies in the given interval. (Round your answer to six decimal places.) |𝑅3 (𝑥)| ≤ 0.000001 Answer: The given function is 𝑓(𝑥) = ln(1 + 2𝑥) . We have to find Tayulor polynomial of degree 3. Note that the formula for the Taylor Polynomial, about the point a, of degree n is given by 𝑇𝑛 (𝑥) = ∑𝑛𝑘=0 𝑓 (𝑘) (𝑎) 𝑘! (𝑥 − 𝑎)𝑘 For 𝑛 = 3 and 𝑎 = 3 we have 𝑃(𝑥) = 𝑇3 (𝑥) = 𝑓(3) + 𝑓 ′(3) 1! (𝑥) + 𝑓 ′′ (3) 2! (𝑥)2 + 𝑓 ′′′ (3) 3! (𝑥)3 Now note the following 𝑓(3) = ln(7) 𝑓 ′ (𝑥) = (ln(2𝑥 + 1))′ = 𝑓 ′′ (𝑥) = ( ′ 2 2𝑥+1 2 2𝑥+1 ⇒ 𝑓 ′ (3) 2 7 4 ) = − (2𝑥+1)2 ⇒ 𝑓 ′′ (3) = − ′ 4 4 49 16 16 𝑓 ′′′ (𝑥) = (− (2𝑥+1)2) = (2𝑥+1)3 ⇒ 𝑓 ′′′ (3) = 343 Plugging above values in the formula for Taylor polynomial we get 𝑃(𝑥) = ln(7) 0! 2 7 0 (𝑥 − (3)) + 1! 1 (𝑥 − (3)) + − 4 49 2! 2 (𝑥 − (3)) + 16 343 3! 3 (𝑥 − (3)) We know that the error in the approximation 𝑓(𝑥) ≈ 𝑇𝑛 (𝑥) is given by 𝑀 𝑅𝑛 (𝑥) = (𝑛+1)! (𝑥 − 𝑎)𝑛+1 where 𝑀 = 𝑚𝑎𝑥𝑧∈[2.6,3.4] = |𝑓 (𝑛+1) (𝑧)| Now notice that 16 ′ 96 𝑓 4 (𝑥) = ((2𝑥+1)3 ) = − (2𝑥+1)4 Therefore we have 96 96 𝑀 = (2×(2.6)+1)4 = (6.2)4 = 0.064968 Thus we get 𝑅3 (𝑥) = 0.064968 4! (𝑥 − 3)4 = 0.002707(𝑥 − 3)4 Thus we get |𝑅3 (𝑥)| < 0.002707 × (0.4)4 ≈ 0.000069 Result: 2 2 7 49 Final answer: 𝑇3 (𝑥) = ln(7) + (𝑥 − 3) − (𝑥 − 3)2 + 8 1029 (𝑥 − 3)3 Question #9 Evaluate the difference quotient for the given function. Simplify your answer. 𝑓(𝑥) = 4 + 3𝑥 − 𝑥 2 , 𝑓(3+ℎ)−𝑓(3) ℎ Answer: 𝑓(𝑥) = 4 + 3𝑥 − 𝑥 2 , therefore 𝑓(3 + ℎ) = 4 + 3(3 + ℎ) − (3 + ℎ)2 = 4 + 9 + 3ℎ − (9 + 6ℎ + ℎ2 ) = 4 − 3ℎ − ℎ2 𝑓(3+ℎ)−𝑓(3) ℎ = ℎ(−3−ℎ) ℎ Result: = (4−3ℎ−ℎ2 )−4 ℎ = −3 − ℎ 𝑓(3+ℎ)−𝑓(3) ℎ = −3 − ℎ Question #10 The joint density function of X and Y is given by 𝑓(𝑥, 𝑦) = 𝑥𝑒 −𝑥(𝑦+1) 𝑥 > 0, 𝑦 > 0 Find the conditional density of Z, given 𝑌 = 𝑦, and that of Y, given 𝑋 = 𝑥. Answer: First of all, let's find marginal PDF's of X and Y. For 𝑥 > 0, we have that ∞ ∞ ∞ ∞ 𝑦=∞ 𝑓𝑋 (𝑥) = ∫0 𝑓(𝑥, 𝑦)𝑑𝑦 = ∫0 𝑥𝑒 −𝑥(𝑦+1) 𝑑𝑦 = −𝑒 −𝑥(𝑦+1)|𝑦=0 = 𝑒 −𝑥 and for 𝑦 > 0 we have that 𝑓𝑌 (𝑦) = ∫0 𝑓(𝑥, 𝑦)𝑑𝑥 = ∫0 𝑥𝑒 −𝑥(𝑦+1) 𝑑𝑥 In order to solve this integral, use integration by parts. Define 𝑢 = 𝑥 and 𝑑𝑣 = 𝑒 −𝑥(𝑦+1) 𝑑𝑥 . Thus 𝑓𝑌 (𝑦) = − 𝑥 𝑦+1 ∞ ⋅ 𝑒 𝑥(𝑦+1)|0 + ∞ 1 ∫ 𝑒 −𝑥(𝑦+1) 𝑑𝑥 𝑦+1 0 1 = (𝑦+1)2 Now, the conditional density of X given 𝑌 = 𝑦 is 𝑓 𝑓(𝑥,𝑦) 𝑥𝑒−𝑥(𝑦+1) = =𝑥𝑒 −𝑥𝑦 𝑓𝑌 (𝑦) 𝑒−𝑥 𝑋|𝑌(𝑥 |𝑦)= Question #11 𝑓(𝑥) = 3𝑥 2 − 2𝑥 a) Verify that 𝑓 ′ (𝑥) = 6𝑥 − 2 b) Find the instantaneous rate of change of f(x) at 𝑥 = −1 Answer: Given: 𝑓(𝑥) = 3𝑥 2 − 2𝑥 a) 𝑓 ′ (𝑥) = 𝑙𝑖𝑚(ℎ→0) (𝑓(𝑥+ℎ)−𝑓(𝑥)) ℎ 2 𝑓(𝑥 + ℎ) = 3(𝑥 + ℎ) − 2(𝑥 + ℎ) = 3(𝑥 2 + 2𝑥ℎ + ℎ2 ) − 2𝑥 − 2ℎ = 3𝑥 2 + 6𝑥ℎ + 3ℎ2 − 2𝑥 − 2ℎ And since 𝑓(𝑥) = 3𝑥 2 − 2𝑥 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = 3𝑥 2 + 6𝑥ℎ + 3ℎ2 − 2𝑥 − 2ℎ − (3𝑥 2 − 2𝑥) = 6𝑥ℎ + 3ℎ2 − 2ℎ; 𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ = 6𝑥ℎ+3ℎ2 −2ℎ ℎ = 6𝑥 + 3ℎ − 2 Now the first equation becomes 𝑓 ′ (𝑥) = lim (6𝑥 + 3ℎ − 2) ℎ→0 = 6𝑥 − 2 b) 𝑓 ′ (−1) = 6(−1) − 2 = −8 Result: a) 𝑓 ′ (𝑥) = 6𝑥 − 2 b) −8 Question #12 How can the graph of 𝑦 = −6√𝑥 can be obtained from the graph of 𝑦 = √𝑥 ? Answer: By vertically stretching by applying a factor of |−6| = 6 (in this instance) and reflecting across the x-axis. Question #13 𝑓(𝑥) = −0.02𝑥 + 11.9 What is the rate of change? Answer: Given: 𝑓(𝑥) = −0.02𝑥 + 11.9 The rate of change for a linear function (𝑦 = 𝑚𝑥 + 𝑏) is the slope m. Looking at the function, 𝑚 = −0.02. So the rate of change is -0.02 gallons per miles. Question #14 Alpinists reach a ledge that is 20 ft above. The height (in feet) is given by the function ℎ = −16𝑡 2 + 32𝑡 + 5 , where t is the time that the hook is in the air (in seconds). Can they throw it high enough to reach the ledge? What time (in seconds) will it take for the hook to reach the maximum height? Answer: ℎ = −16𝑡 2 + 32𝑡 + 5 𝑎 < 0, so the maximum occurs at: − 𝑏 2𝑎 , 𝑓 (− 𝑏 2𝑎 ). Substitute to find what is the t for the maximum: 𝑡=− 32 2(−16) =− 32 −32 =1 Substitute for maximum height: ℎ(1) = −16(1)2 + 32(1) + 5 = −16 + 32 + 5 = 16 + 5 = 21 ft It will 1 second for the hook to reach its max height, which is 21ft. Since the ledge is 20 ft high and 21>20, they can throw it high enough. Question #15 Classify each function: 𝑥–7 a) 𝑦= b) 𝑦=𝑥+ c) 𝑦 = 9𝑥 𝑥+7 𝑥2 √𝑥–3 d) e) f) 𝑦 = 𝑥9 𝑦 = 2𝑡 6 + 𝑡 4 − y = cos  + sin  Answer: a) 𝑦 = 𝑥–7 – rational function 𝑥+7 b) 𝑦 = 𝑥 + 𝑥2 √𝑥–3 – root function c) 𝑦 = 9𝑥 – logaritmic function d) 𝑦 = 𝑥 9 – power function e) 𝑦 = 2𝑡 6 + 𝑡 4 – – polynomial function of degree 6 f) y = cos  + sin  - trigonometric function Question #16 Please, classify following functions: a) 𝑓(𝑥) = log 3 (𝑥) 3 b) 𝑔(𝑥) = √𝑥 2𝑥 3 c) ℎ(𝑥) = d) e) 𝑢(𝑡) = 1– 1.1𝑡2.53 𝑡 2 𝑣(𝑡) = 6𝑡 f) w( ) = sin cos 2 5–𝑥 2 Answer: a) 𝑓(𝑥) = log 3 (𝑥) – logarithmic function 3 b) 𝑔(𝑥) = √𝑥 – root function c) ℎ(𝑥) = 2𝑥 3 5–𝑥 2 – rational function d) 𝑢(𝑡) = 1– 1.1𝑡2.53 𝑡 2 – polynomial function e) 𝑣(𝑡) = 6𝑡 power function f) w( ) = sin cos  2 – trigonometric function Question #17 Classify each function, please: a) b) 𝑓(𝑥) = log 7 (𝑥) 8 𝑔(𝑥) = √𝑥 c) ℎ(𝑥) = d) e) 𝑢(𝑡) = 1– 1.1𝑡2.57 𝑡 2 𝑣(𝑡) = 6𝑡 f) w( ) = sin  cos5  2𝑥 3 8–𝑥 2 Answer: a) 𝑓(𝑥) = log 7 (𝑥) – logarithmic function 8 b) 𝑔(𝑥) = √𝑥 – root function c) ℎ(𝑥) = 2𝑥 3 8–𝑥 2 – rational function d) 𝑢(𝑡) = 1– 1.1𝑡2.57 𝑡 2 – polynomial function e) 𝑣(𝑡) = 6𝑡 – power function f) w( ) = sin  cos 5  – trigonometric function Question #18 Classify each function: a) b) 𝑓(𝑥) = log 4 (𝑥) 5 𝑔(𝑥) = √𝑥 c) ℎ(𝑥) = d) e) 𝑢(𝑡) = 1– 1.1𝑡2.52 𝑡 2 𝑣(𝑡) = 3𝑡 f) w( ) = sin  cos 7  2𝑥 3 9–𝑥 2 Answer: a) 𝑓(𝑥) = log 4 (𝑥) – logarithmic function 5 b) 𝑔(𝑥) = √𝑥 – root function c) ℎ(𝑥) = 2𝑥 3 9–𝑥 2 – rational function d) 𝑢(𝑡) = 1– 1.1𝑡2.52 𝑡 2 – polynomial function e) 𝑣(𝑡) = 3𝑡 – power function f) w( ) = sin  cos 7  – trigonometric function Question #19 Classify each function: a) b) 𝑓(𝑥) = log 3 (𝑥) 5 𝑔(𝑥) = √𝑥 c) ℎ(𝑥) = d) e) 𝑢(𝑡) = 1– 1.1𝑡2.51 𝑡 2 𝑣(𝑡) = 3𝑡 f) w( ) = sin  cos 6  2𝑥 3 8–𝑥 2 Answer: a) 𝑓(𝑥) = log 3 (𝑥) – logarithmic function 5 b) 𝑔(𝑥) = √𝑥 – root function c) ℎ(𝑥) = 2𝑥 3 8–𝑥 2 – rational function d) 𝑢(𝑡) = 1– 1.1𝑡2.51 𝑡 2 – polynomial function e) 𝑣(𝑡) = 3𝑡 – power function f) w( ) = sin  cos 6  – trigonometric function Question #20 Classify, please, these functions: 𝑥–9 a) 𝑦= b) 𝑦=𝑥+ 𝑥+9 𝑥2 √𝑥–2 Answer: a) 𝑦 = b) 𝑦 = 𝑥–9 𝑥+9 𝑥–9 𝑥+9 – rational function – root function