Functions #9 - Questions and Answers

Question 1 Write y = 2x + 7 using function notation. Answer Equation y = 2x + 7 Function notation f(x) = 2x + 7 Question 2 For what value of the constant c is the function f continuous on (-∞, ∞)? f(x) = { (cx² + 2x) / (x² - 1) if x < 2; x³ - cx if x ≥ 2 } Answer 1 f(x) = { (cx² + 2x) / (x² - 1) if x < 2; x³ - cx if x ≥ 2 } Note that: cx²+2x and x³-cx are polynomials and are always continuous. We only need to check for the continuity at x = 2 f(x) will be continuous at x = 2, if and only if lim(x→2-) f(x) = lim(x→2+) f(x) lim(x→2-) (cx² + 2x) / (x² - 1) = lim(x→2+) x³ - cx c * 2² + 2*2 = 2³ - c * 2 4c + 4 = 8 - 2c 6c = 4 c = 2/3 Result: c = 2/3 Answer 2 For the function f(x) to be continuous on (-∞, ∞), we need to ensure that the left- hand and right-hand pieces of the function match at x = 2. This means that the limit of f(x) as x approaches 2 from both sides should be equal. First, let's consider the left-hand piece of the function for x < 2: f(x) = (cx² + 2x) / (x² - 1) To find the limit as x approaches 2 from the left, we substitute x = 2 into the expression: lim(x→2-) f(x) = (c(2²) + 2(2)) / (2² - 1) Simplifying further: lim(x→2-) f(x) = (4c + 4) / 3 Next, let's consider the right-hand piece of the function for x ≥ 2: f(x) = x³ - cx To find the value of f(x) at x = 2, we substitute x = 2 into the expression: f(2) = 2³ - c(2) = 8 - 2c To ensure continuity at x = 2, the limit and the function value must be equal. Therefore, we have: lim(x→2-) f(x) = f(2) Substituting the expressions we found earlier: (4c + 4) / 3 = 8 - 2c To solve this equation for c, we can multiply both sides by 3 to eliminate the fraction: 4c + 4 = 24 - 6c Combining like terms: 10c = 20 Dividing both sides by 10: c = 20/10 = 2 Hence, the constant c that makes the function f(x) continuous on (-∞, ∞) is c = 2 Answer 3 To find the value of the constant (c) for which the function (f) is continuous on (-∞,∞), we need to ensure that the left-hand limit and right-hand limit of (f(x)) as (x) approaches 2 from both sides are equal. First, let's calculate the left-hand limit of (f(x)) as (x) approaches 2: lim(x→2-) f(x) = lim(x→2-) (cx² + 2x) / (x² - 1) Substituting (x = 2) into the expression, we have: lim(x→2-) f(x) = (c(2)² + 2(2)) / (2)² - 1 = (4c + 4) / 3 Now, let's calculate the right-hand limit of (f(x)) as (x) approaches 2: lim(x→2+) f(x) = lim(x→2+) (x³ - cx) Substituting (x = 2) into the expression, we have: lim(x→2+) f(x) = (2)³ - c(2) = 8 - 2c For the function (f(x)) to be continuous at (x = 2), the left-hand limit and right-hand limit must be equal. Therefore, we have the equation: (4c + 4) / 3 = 8 - 2c To solve for (c), we can start by multiplying both sides of the equation by 3 to eliminate the fraction: 4c + 4 = 3(8 - 2c) Next, distribute 3 on the right-hand side: 4c + 4 = 24 - 6c Now, let's simplify the equation by combining like terms: 4c + 6c = 24 - 4 10c = 20 Finally, divide both sides of the equation by 10 to solve for (c): c = 20/10 = 2 Hence, the value of the constant (c) for which the function (f) is continuous on (-∞, ∞) is (c = 2). Question 3 Whether each of these functions is a bijection from R to R. a) f(x) = -3x + 4 b) f(x) = -3x² + 7 c) f(x) = (x + 1) / (x + 2) d) f(x) = x³ + 1 Answer Part a- f(x) = -3x + 4 is a bijection. It is one-to-one as f(x) = f(y) => -3x + 4 = -3y + 4 => -3x = -3y => x = y It is onto as f( (4 - x) / 3 ) = x Part b- f(-x) = f(x). The function is not a bijection as a result. Part c- No real number exists that is such that f(x) = (x + 1) / (x + 2) = 1. Hence the function is not a bijection. Part d- f(x) = x³ + 1 is a bijection. It is a strictly increasing function. Result: Part a - Yes Part b - NO Part c - No Part d - Yes Question 4 Find two numbers whose difference is 100 and whose product is a minimum. Answer It is given that the difference of two numbers is 100. x - smaller y - larger y - x = 100 y = 100 + x We write a function that represents the minimum product of two numbers. f(x,y) = xy Substitute y = 100 + x into f(x, y) = xy. f(x) = x(100 + x) = 100x + x² Now, we find the first derivative of the function f(x) = 100x + x². f'(x) = (100x + x²)' = (100x)' + (x²)' = 100 + 2x Find the critical points. Solve the equation f'(x) = 0. f'(x) = 100 + 2x 0 = 100 + 2x -100 = 2x x = -100/2 = -50 Now, find the second derivative of the function f(x). Use the second derivative test to determine whether the number we found was a critical number. f''(x) = (100 + 2x)' = (100)' + (2x)' = 0 + 2 = 2 > 0 Since, it is positive, it means that yes, there is a minimum. Find y. Substitute x = -50 into y - x = 100. y + 50 = 100 y = 50 Results: x = -50, y = 50 Question 5 Determine the missing coordinates of the points on the graph of the function. y = arctan x Answer y = arctan(-√3) (-√3,y) y = -π/3 (-√3, -π/3) π/6 = arctan(x) (x, π/6) tan(π/6) = x x = √3/3 (√3/3, π/6) π/4 = arctan(x) (x, π/4) tan(π/4) = x x= 1 (1, π/4) Results: (from left to right) (-√3, -π/3), (√3/3, π/6), (1, π/4)