Functions #1 - Questions and Answers

Question 1 Find the constant a such that the function is continuous on the entire real number line. f(x) = { x3 x ≤ 2 ax2 x > 2 Answer 1 f(x) = { x3 x<2 ax2 x > 2 Given that f is continuous down the complete real number line, x=2 is also where f is continuous. This means lim f(x) = lim f(x) = lim f(x) = f(2) →2+ →2- →2 ... lim f(x) = ax2 = 4a →2+ ... lim f(x) = x3 = 23 = 8 →2- ... lim f(x) = 8 z→2+ ... 4a = 8 a = 2 Result: a = 2. Answer 2 Result: The constant a can be any real number. Solution: To find the constant a such that the function f(x) is continuous on the entire real number line, we need to ensure that the two cases of the function match up at the point x = 2. Let's analyze the left-hand side of the function, x³, for x ≤ 2. Since x³ is a polynomial, it is continuous for all real values of x. Now, let's consider the right-hand side of the function, ax², for x > 2. To ensure continuity at x = 2, the limit of ax² as x approaches 2 from the left should be equal to the value of ax² at x = 2. To express this mathematically, we can write: limx→2- ax2 = ax2|x=2 Evaluating the limit gives: limx→2- ax2 = a(2)2 = 4a And the value of ax2 at x = 2 is: ax2|x=2 = a(2)2 = 4a Therefore, we need to find the value of a that satisfies the equation: limx→2- ax2 = ax2|x=2 Substituting the expressions we derived earlier, we have: 4a = 4a In conclusion, the constant a can be any real number. Answer 3 Step 1: Let's evaluate f(x) at x = 2 in both cases: For x ≤ 2, we have: f(x) = x³ Substituting x = 2, we get: f(2) = 2³ = 8 For x > 2, we have: f(x) = ax2 Substituting x = 2, we get: f(2) = a(2)² = 4a Step 2: To ensure continuity at x = 2, the two expressions for f(2) must be equal. Therefore, we have: 8 = 4a Solving this equation for a, we find: a = 8/4 = 2 Hence, the constant a that makes the function f(x) continuous on the entire real number line is a = 2. Answer 4 To find the constant a such that the function f(x) is continuous on the entire real number line, we need to ensure that the left-hand limit of f(x) as x approaches 2 is equal to the right-hand limit of f(x) at x = 2, and they are both equal to the value of f(x) at x = 2. Let's calculate these limits and evaluate the function at x = 2 to find the value of a. The left-hand limit of f(x) as x approaches 2 is given by: limx→2- f(x) =limx→2- x³ Since x³ is a continuous function, we can directly substitute x = 2 into the function: limx→2- f(x) = 2³ = 8 The right-hand limit of f(x) as x approaches 2 is given by: limx→2+ f(x) =limx→2+ ax2 Again, we can substitute x = 2 into the function: limx→2+ f(x) = a·22 = 4a To ensure continuity, we set these two limits equal to each other: 8 = 4a Now, we can solve for a by dividing both sides of the equation by 4: 8/4 = 4a/4 Simplifying, we have: 2 = a Hence, the constant a that makes the function f(x) continuous on the entire real number line is 2. Question 2 Find the nth Maclaurin polynomial for the function. f(x) = e−x, n = 5 Answer Given Polynomial f(x) = e−x Then nth Maclaurin polynomial is f(x) = f(0) + f'(0)x + f''(0)x2/2!+ f'''(0)x3/3! +...+ fⁿ(0)xⁿ/n! Then f(0) = e−⁰ = 1 f'(0) = −1e−⁰ = -1 f''(0) = e−⁰ = 1 f'''(0) = -e−⁰ = -1 f''''(0) = e−⁰ = 1 f'''''(0) = -1e−⁰ = -1 Therefore Maclaurin polynomial for n=5 is f(x) = 1 - x + x2/2! - x3/3! + x4/4! - x5/5! Result: f(x) = 1 - x + x²/2! - x³/3! + x⁴/4! - x⁵/5! Question 3 Find the exact length of the curve. y= x³/3 + 1/(4x), 1 ≤ x ≤ 2 Answer The function f(x) = x³/3 + 1/(4x) is continuous, differentiable on [1,2] and its derivative f'(x) = x2 - 1/(4x²) is continuous on [1,2] so the length L of the curve is given by the formula: L = ∫[1,2] √(1 + (x² - 1/(4x²))²) dx therefore, L = ∫[1,2] √(1 + (x⁴ - 1/2 + 1/(16x⁴))) dx = ∫[1,2] √((x⁴ + 1/2 + 1/(16x⁴))) dx = ∫[1,2] √(x² + 1/(4x²))² dx = ∫[1,2] (x² + 1/(4x²)) dx =[x³/3 - 1/(4x)] [1,2] = [(2³/3 - 1/(42))]-[(1³/3 - 1/(41))] = (8/3 - 1/8) - (1/3 - 1/4) = 64/24 - 3/24 - 8/24 + 6/24 = 59/24 L=59/24 Question 4 Find the absolute maximum and minimum values of f on the given interval. f(x) = 4x³- 6x² - 24x + 9.[-2,3] Answer Step 1 We will find the first derivative using the power rule f(x) = 4x³- 6x² - 24x + 9 f'(x) = 4(3x²) -6(2x) - 24 f'(x) = 12x² - 12x - 24 Step 2 We find the critical values by solving f'(x) = 0 12x²-12x - 24 = 0 12(x²- x - 2) = 0 12(x-2)(x + 1) = 0 x = 2, -1 Then we find the values of the function at the critical points and at the endpoints. f(x) = 4x³-6x² - 24x + 9 f(-2) = 4(-2)³-6(-2)²-24(-2) + 9 = 1 f(-1) = 4(-1)³-6(-1)²-24(-1) + 9 = 23 (max) f(2)= 4(2)³-6(2)² - 24(2) + 9 = −31 (min) f(3) = 4(3)³-6(3)² - 24(3) + 9 = -9 Answer: Absolute minimum value= -31 Absolute maximum value=23 Question 5 A baseball team plays in a stadium that holds 55,000 spectators. With ticket prices at 10, the average attendance had been 27,000. When ticket prices were lowered to10,the average attend ance had been 27,000.When ticket prices were lowered to 8, the average attendance rose to 33,000. How should ticket prices be set to maximize revenue? Answer a) p(x) = -1/3000x + 19 R(x) = x*p(x) = -1/3000x² + 19x R'(x) = 0 = -2/3000x + 19 0 = -1/1500x + 19 1/1500x= 19 x= 28500 Since R'' = -1/1500 < 0, R is concave down here and this is a maximum p(28500) = -1/3000(28500) +19 = $9.50 Answer is $9.50