Analyzing functions #1 - Questions and Answers

Question 1 When should I use brackets or parenthesis in finding domain or range? Answer 1 You should use a bracket (square bracket) to indicate that the endpoint is included in the interval, and a parenthesis (round bracket) to indicate that it is not. Brackets are like inequalities that say "or equal" and parentheses are like strict inequalities. For example, (3,7) includes 3.1 and 3.007 and 3.00000000002, but it does not include 3. It also includes numbers greater than 3 and less than 7, but it does not include 7. We can say say this is 3 to 7 "exclusive" (Excluding the endpoints) [4,9] includes 4 and every number from 4 up to 9, and it also includes 9. We can say this is 4 to 9 "inclusive" (Including the endpoints) (a, b) = {x: a < x and x < b} [a, b] = {x: a ≤ x and x ≤ b} Of course, there can be also mixed intervals (a,b] or [a,b). The symbols -∞, and ∞ are used to indicate that there is no left or right endpoint for the interval. They always take parentheses. For example: Domain of f(x) = √x is [0,∞), √0 = 0 is a number. Domain of g(x) = 1/√x is (0,∞), 1/√0 is a not number. Answer 2 When determining the domain or range of a function, it is important to use the appropriate notation to clearly express your answer. In mathematics, we use brackets [] and parentheses () to represent intervals. The choice between brackets and parentheses depends on whether the endpoints are included or excluded from the interval. Domain: Use parentheses () when the endpoints are excluded from the interval. Use brackets [] when the endpoints are included in the interval. For example, consider the function f(x) = 1/x: To find the domain of f(x), we need to determine the values of x for which f(x) is defined. Since division by zero is undefined, we exclude x = 0 from the domain. Therefore, the domain of f(x) is (-∞, 0) u (0,∞) or (-∞, 0) u (0,∞) using parentheses. Range: Use parentheses () when the endpoints are excluded from the interval. Use brackets [] when the endpoints are included in the interval. For example, let's consider the function g(x) = x²: To find the range of g(x), we need to determine the set of all possible values that g(x) can take. Since x² is always non-negative (or zero), the range of g(x) is [0,∞) or [0,∞) using brackets. In summary, parentheses () are used to exclude endpoints, while brackets [] are used to include endpoints. It is important to use the appropriate notation to accurately represent the domain or range of a function. Answer 3 The choice between using brackets or parentheses when finding the domain or range depends on whether the endpoints of the interval are included or excluded. Brackets [] are used to denote that the endpoint is included in the interval. For example: The domain of a function defined on a closed interval [a, b] would be written as {x ∈ R: a ≤ x ≤ b}. The range of a function defined on a closed interval [c, d] would be written as {y ∈ R: c ≤ y ≤ d}. Parentheses () are used to indicate that the endpoint is excluded from the interval. For example: The domain of a function defined on an open interval (a, b) would be written as {x ∈ R: a < x < b}. The range of a function defined on an open interval (c, d) would be written as {y ∈ R: c < y < d}. It's important to consider the context and requirements of the problem to determine whether to use brackets or parentheses. Question 2 At what point does the curve have maximum curvature? What happens to the curvature as x tends to infinity y = ln x Answer f(x) = ln x f'(x) = 1/x f''(x) = -1/x² Plug into the curvature formula k(x) = |f"(x)| / (1 + (f'(x))²)³⁄² = |-1/x²| / (1 + (1/x)²)³⁄² Find k' k(x) = 1/x² * [1 + 1/x²]⁻³⁄² = x⁻²[1 + x⁻²]⁻³⁄² k'(x) = -2x⁻³[1 + x⁻²]⁻³⁄² + x⁻² * (-3/2)[1 + x⁻²]⁻⁵⁄² * (-2x⁻³) = -2x⁻³[1 + 1/x²]⁻³⁄² + 3x⁻⁵[1 + 1/x²]⁻⁵⁄² = -2/x³(1 + 1/x²)⁻³⁄² + 3/x⁵(1 + 1/x²)⁻⁵⁄² = (-2x²/x⁵(1 + 1/x²)³⁄² + 3/x⁵(1 + 1/x²)⁵⁄²) = (-2x² + 3)/(x⁵[1 + 1/x²]⁵⁄²) Find where k'=0 0 = 3 - 2x² 2x² = 3 x = ±√3/2 ≈ ±0.7071 The negative root is not in the domain of ln x. We can check numbers in the surrounding intervals to make sure it is the max: (0, √3/2): k′(0.1) ≈ 0.96 (√3/2, ∞): k′(1) ≈ -0.18 k is increasing then decreasing, so it is the max. As x → ∞: lim (1/x²)/ (1 + 1/x²)³⁄² = 0 x->∞ the curvature approaches 0. Question 3 Which of the following graphs represent functions that have inverse functions? [Image of 3 graphs labeled a, b, and c] Answer Step 1: Horizontal line test: Horizontal line test is used to determine whether a function has inverse or not. We draw the graph of given function and check is there any horizontal line that can intersect the graph in more than 1 point. If no horizontal line intersects graph in more than 1 point, we say, the function has an inverse. If any horizontal line intersect the graph in more than 1 point, the function does not have an inverse. Let us check given graphs one by one Step 2: (a) For the given graph, [Graph of function a on a coordinate plane] if we draw horizontal line, it intersects the graph in two distinct points So horizontal line test fails. Hence, function represented by given graph does not have an inverse function. Step 3: (b) For the given graph, [Graph of function b on a coordinate plane] if we draw any horizontal line, it intersects the graph in only one point. So the graph passes the horizontal line test. Hence, function represented by given graph has an inverse function. Step 4: (c) For the given graph, [Graph of function c on a coordinate plane] if we draw any horizontal line, it intersects the graph in only one point. So the graph passes the horizontal line test. Hence, function represented by given graph has an inverse function. Question 4 If f(2) = 10 and f'(x) = x²f(x) for all x, find f"(2) Answer 1 Step 1 It is given that f'(x) = x²f(x) Substitute x = 2 f'(2) = 2² x f(2) Substitute f(2) = 10 f'(2) = 4 × 10 = 40 f'(x) = x²f(x) Differentiate using the product rule f''(x) = (x²)' × f(x) + x² × f'(x) f''(x) = (2x) × f(x) + x² × f'(x) Substitute x = 2 f''(2) = 2 × 2 × f(2) + 2² × f'(2) Substitute f(2) = 10 and f'(2) = 40 f''(2) = 4 x 10 + 4 x 40 = 40 + 160 = 200 Answer 2 Step 1 Metod 1: Divide by f(x) f'(x)/f(x) = x² Step 2 Differentiate both sides (f''(x)f(x) - f'(x)²)/f(x)² = 2x Step 3 At x = 2 f'(2) = 2²f(2) f'(2) = 4 × 10 = 40 (f''(2)f(2) - f'(2)²)/f(2)² = 2×2 (10f''(2) - 40²)/10² = 4 10f''(2) - 1600 = 400 10f''(2) = 400 + 1600 Step 4 Solve for f"(x) f''(2) = 2000/10 = 200 Step 5 Method 2 (Rose Winter): Use the product rule f''(x) = 2x f(x) + x²f'(x) Step 6 As before, at x = 2 we can calculate: f'(2) = 2²f(2) f'(2) = 4 × 10 = 40 Step 7 Substituting the values we get: f''(2) = 2 x 2 x 10 + 2² x 40 = 40 + 160 = 200 Answer 3 Step 1: Let's begin by finding the first derivative of the function f(x). We are given that f'(x) = x²f(x). f'(x) = x²f(x) Next, we need to find the second derivative f"(x). To do this, we differentiate the expression f'(x) with respect to x. f"(x) = d/dx(x²f(x)) Step 2: Now, let's substitute x = 2 into the given information f(2) = 10 to find the value of f(2). f(2) = 10 Finally, we can evaluate f"(2) by substituting x = 2 into the expression we obtained for f''(x). f"(2) = d/dx(x²f(x))|x=2 Answer 4 Result: f''(2) = 200 Solution: We are given that f(2) = 10 and f'(x) = x²f(x) for all x. We need to find f"(2). To find f"(2), we can differentiate the given expression f'(x) = x²f(x) with respect to x. Differentiating both sides of the equation, we have d/dx [f'(x)] = d/dx [x²f(x)] Using the chain rule, the left side becomes f"(x) and the right side becomes 2xf(x) + x²f'(x). Therefore, we have f"(x) = 2xf(x) + x²f'(x). Now, substituting x = 2 into the equation, we get f''(2) = 22f(2) + 2²f'(2) Substituting the given values f(2) = 10 and f'(2) = 2²f(2) = 410 = 40, we can calculate f''(2) as f''(2) = 2210 + 2²40 Simplifying the expression, we find f''(2) = 40 + 4*40 f''(2) = 40 + 160 f''(2) = 200 Therefore, f"(2) = 200. Answer 5 Given information: f(2) = 10 f'(x) = x²f(x) To find f''(2), we can differentiate f'(x) with respect to x and then substitute x = 2. Let's start by finding f''(x) using the chain rule: d/dx [f'(x)] = d/dx [x²f(x)] To apply the chain rule, we differentiate f(x) with respect to x while treating x² as a constant: f''(x) = 2xf(x) + x²f'(x) Now, let's substitute x = 2 into the above equation to find f"(2): f"(2) = 2(2)f(2) + (2)²f'(2) Using the given information that f(2) = 10, we can substitute the values into the equation: f"(2) = 2(2)(10) + (2)²f'(2) Simplifying further: f"(2) = 40 + 4f'(2) We still need the value of f'(2). Let's differentiate f'(x) using the given information that f'(x) = x²f(x): d/dx [f'(x)] = d/dx[x²f(x)] Applying the quotient rule: f'(x) = (x²f'(x) - 2xf(x)) We can simplify this equation by multiplying through by x^4: x⁴f''(x) = x²f'(x) - 2xf(x) Rearranging the terms: x⁴f''(x) - x²f'(x) + 2xf(x) = 0 Now, we can substitute x = 2 into the above equation to find f'(2): (2)⁴f''(2) - (2)²f'(2) + 2(2)f(2) = 0 Simplifying further: 16f''(2) - 4f'(2) + 4f(2) = 0 12f''(2) + 4f(2) = 0 Now, substitute f(2) = 10 into the equation: 12f'(2) + 4(10) = 0 12f'(2) + 40 = 0 12f'(2) = -40 Finally, we can substitute the value of f'(2) back into the equation for f''(2): f''(2) = 40 + 4f'(2) f"(2) = 40 + 4(-40/12) f''(2) = 40 - 160/3 = -120/3 f"(2) = -120/3 Therefore, f"(2) = -120. Question 5 Is the statement true or false? "The domain and the range of the reciprocal function are the set of all real numbers." Answer 1 False. The domain and range of the reciprocal function are all real numbers except the number 0 since f(0) = 1/0 is undefined. If x can't be 0, y can't be 0 either. [graph of f(x) = 1/x, with x and y axes labeled] Answer 2 The reciprocal function is f(x) = 1/x. The domain of f(x) is the set {x|x ≠0}. Therefore, the statement is False