Functions #3 - Questions and Answers

Question 1 Plot the complex number and find its absolute value. z=9+7i Answer Concept used: The absolute value or modulus of complex number is the distance between the origin (0, 0) and point (a,b) in the complex plane. Formula to find the absolute value of any complex number: |z| = √(x² + y²) or |z| = √(a² + b²). Calculation: The complex number is Interprets as z=a+bi can be represent as the point (a,b) in the complex plane. The complex number z=9+7i can be represented as the point (9,7) in the complex plane. The plot complex number z=9+7i as shown in below: [Image of a graph with x axis labeled real axis and y axis labeled imaginary axis. The point (9,7) is plotted on the graph and a line drawn from (0,0) to (9,7)] The absolute value of z=a+bi is |a + bi| = √(a² + b²). |z| = √(9² + 7²) => √81 + 49 = √130 Hence, the absolute value is √130. Question 2 Match the function with its name: f(x) = c a) squaring function b) square root function c) cubic function d) linear function e) constant function f) absolute value function g) greatest integer function h) reciprocal function i) identity function Answer The function f(x) = c is called a constant function. Its domain is independent of the input values, while its range is fixed at "c". Thus, the correct answer is (e). Question 3 Find a function f such that f'(x) = 5x³ and the line 40x + y = 0 is tangent to the graph of f. Answer Step 1 We have to find a function f such that f'(x) = 5x³ and the line 40x + y = 0 is tangent to the graph of f First, notice that by integrating f'(x) = 5x³ we get f(x) = 5x⁴/4 + C where C is a integrating constant. We have to find the value of C. Now notice that the slope of any tangent to the graph of f(x) is given by f'(x) Also, note that the slope of the line y = -40x is given by -40 Thus in order to the line 40x + y = 0 become a tangent to the graph of f(x) We must have a value for x that allows for f'(x) = -40 Thus we have solve 5x³ = -40 This provides x³ = -8 => x = -2 [Considering only real cubic roots of -8] Thus, at x=-2, the line 40x + y = 0 become a tangent to the graph of f(x) Step 2 As x = -2, f(x) touch the line y = -40x hence, we achieve f(-2) = -40*(-2) = 80 Again pluggin x=-2 on the expression f(x) = 5x⁴/4 + C, we get 80 = 5*(-2)⁴/4 + C => C = 80 - 20 = 60 Therefore, the necessary function is f(x) = 5x⁴/4 + 60 Answer f(x) = 5x⁴/4 + 60 Question 4 Let's say that f and g are continuous functions, and g(2) = 6 and lim [3f(x) + f(x)g(x)] = 36 Find f(2), x → 2 Answer 1 Let h(x) = 3f(x) + f(x)g(x) Since f(x) and g(x) are continuous, by theorem 4 h(x) is continuous That means lim h(x) = h(2) →2 lim 3f(x) + f(x) g(x) = 3f(2) + f(2) g(2) →2 Keep in mind: Given that g(2) = 6 and the maximum in the RHS is 36 36 = 3f(2) + f(2).6 36 = 9f(2) 4 = f(2) Result: f(2) = 4 Answer 2 Explanation: We know that g(2) = 6 and limx→2 (3f(x) + f(x)g(x)) = 36. We're asked to find f(2). Let's start by manipulating the limit expression using algebraic operations: limx→2 (3f(x) + f(x)g(x)) = 36 We can factor out f(x) from both terms: limx→2 (f(x)(3 + g(x))) = 36 Since the limit is taken as x approaches 2, we can substitute x = 2 into the expression: f(2)(3 + g(2)) = 36 Substituting the known value g(2) = 6: f(2)(3 + 6) = 36 Simplifying further: 9f(2) = 36 Now, we can solve for f(2) by dividing both sides of the equation by 9: f(2) = 36/9 = 4 Therefore, the value of f(2) is 4. Answer 3 Given that f and g are continuous functions, and g(2) = 6, we are given that limx→2 (3f(x) + f(x)g(x)) = 36. We need to find the value of f(2). Let's start by manipulating the limit expression using algebraic operations. We can rewrite the expression as: limx→2 (f(x)(3 + g(x))) = 36 Since the limit is taken as x approaches 2, we can substitute x = 2 into the expression: f(2)(3 + g(2)) = 36 Substituting the given value g(2) = 6: f(2)(3 + 6) = 36 Simplifying further: f(2)(9) = 36 Dividing both sides of the equation by 9: Dividing both sides of the equation by 9: f(2) = 36/9 = 4 Therefore, the value of f(2) is 4. Answer 4 We are given that f and g are continuous functions. Additionally, g(2) = 6. We need to find f(2) given the limit: limx→2 (3f(x) + f(x)g(x)) = 36 Since we are interested in the behavior as x approaches 2, we will rewrite the limit expression in terms of x approaching 2: limx→2 (3f(x) + f(x)g(x)) = limx→2 f(x)(3 + g(x)) = 36 We can now use the limit definition of continuity to evaluate this expression. We know that g(2) = 6, so as x approaches 2, g(x) approaches 6. Therefore, we have: limx→2 f(x)(3 + g(x)) = limx→2 f(x) (3 + 6) = limx→2 9f(x) = 36 Now, we can simplify the expression: limx→2 9f(x) = 36 Dividing both sides by 9: limx→2 f(x) = 36/9 = 4 Using the limit definition of continuity, we can conclude that f(2) = 4. Therefore, the solution to the problem is f(2) = 4. Question 5 Evaluate the difference quotient for the given function. f(x) = 4 + 3x - x², (f(3+h)-f(3))/h Answer 1 Solution: f(x) = 4 + 3x - x², thus 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 = (4-3h-h²-4)/h = (-3h - h²)/h = h(-3 - h)/h = -3 - h Answer: (f(3+h)-f(3))/h = -3 - h Answer 2 First, let's find f(3 + h): f(3+h) = 4 + 3(3+h) - (3 + h)² Simplifying this expression: f(3+h) = 4 + 9 + 3h - (9 + 6h + h²) f(3+h) = 13 + 3h - 9 - 6h - h² f(3+h) = 4 - 3h - h² Now, let's find f(3): f(3) = 4 + 3(3) - 3² Simplifying this expression: f(3) = 4 + 9 - 9 f(3) = 4 Finally, we can calculate the difference quotient: (f(3+h)-f(3))/h = (4-3h-h²)-4/h Simplifying this expression: (f(3+h)-f(3))/h = (-3h - h²)/h (f(3+h)-f(3))/h = h(-3-h)/h (f(3+h)-f(3))/h = -3 - h Therefore, the difference quotient for the given function is -3 - h. Answer 3 Solution: Substitute the value of x in the function with 3 + h: f(3+h) = 4 + 3(3 + h) - (3 + h)² Evaluate f(3) by substituting x with 3 in the function: f(3) = 4 + 3(3) - (3)² Plug in the values into the difference quotient formula: (f(3+h)-f(3))/h = (4+3(3+h)-(3+h)²)-(4+3(3)-(3)²)/h Answer 4 Step 1: Substitute the values into the function. We need to find (f(3+h)-f(3))/h. Let's substitute these values into the function: f(3+ h) = 4 + 3(3+ h)- (3 + h)² and f(3) = 4 + 3(3) - (3)². Step 2: Simplify the expressions. Now, we simplify the expressions inside the function: f(3+h) = 4 + 9 + 3h - (9 + 6h + h²) and f(3) = 4 + 9 - 9. Step 3: Expand and combine like terms. Expand and simplify the expressions further: f(3+h) = 13 + 3h - 9 - 6h - h² and f(3) = 4. Step 4: Evaluate the difference quotient. Now, let's substitute these values back into the difference quotient: (f(3+h)-f(3))/h = (13 + 3h - 9 - 6h - h² - 4)/h Step 5: Simplify the expression. Simplify the expression by combining like terms: (13 + 3h - 9 - 6h - h² - 4)/h = (0 - 3h - h²)/h Step 6: Further simplify and factor out a negative sign: (-3h - h²)/h = -(h² + 3h)/h Step 7: Cancel out common factors. We can cancel out the common factor of h: -(h(h + 3))/h = -h - 3 Therefore, the difference quotient for the given function f(x) = 4 + 3x - x² is -h - 3, which can also be written as -3 - h.