This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions to the problems: Given functions: f(x) = 2x + 3 where x [-1, 2) g(x) = x^2 with domain x R 2.3.1 Determine the value of g(sqrt(3)). Step 1: Substitute sqrt(3) into the function g(x). g(x) = x^2 g(sqrt(3)) = (sqrt(3))^2 Step 2: Simplify the expression. g(sqrt(3)) = 3 The value of g(sqrt(3)) is 3. 2.3.2 Determine the equation of f^-1(x) in the form y=.... Step 1: Replace f(x) with y. y = 2x + 3 Step 2: Swap x and y to find the inverse. x = 2y + 3 Step 3: Solve for y. x - 3 = 2y y = (x - 3)/(2) The equation of f^-1(x) is y = (x - 3)/(2). 2.3.3 Determine the value of a if g(a+2) - f(a+2) = 0. Step 1: Substitute (a+2) into g(x) and f(x). g(a+2) = (a+2)^2 f(a+2) = 2(a+2) + 3 Step 2: Set up the equation g(a+2) - f(a+2) = 0. (a+2)^2 - (2(a+2) + 3) = 0 Step 3: Expand and simplify the equation. (a^2 + 4a + 4) - (2a + 4 + 3) = 0 a^2 + 4a + 4 - 2a - 7 = 0 a^2 + 2a - 3 = 0 Step 4: Factor the quadratic equation. (a+3)(a-1) = 0 Step 5: Solve for a. a+3 = 0 a = -3 a-1 = 0 a = 1 The values of a are -3 or 1. 2.3.4 Write down the domain for f^-1(x). Step 1: Determine the range of f(x). The domain of f^-1(x) is the range of f(x). The domain of f(x) is x [-1, 2). For f(x) = 2x+3: When x = -1, f(-1) = 2(-1) + 3 = -2 + 3 = 1. When x = 2, f(2) = 2(2) + 3 = 4 + 3 = 7. Since f(x) is a linear function with a positive slope, its range is [1, 7). Step 2: State the domain of f^-1(x). The domain for f^-1(x) is x [1, 7). 2.3.5 Prove that f(f^-1(x)) = f^-1(f(x)). Step 1: Calculate f(f^-1(x)). We have f(x) = 2x+3 and f^-1(x) = (x-3)/(2). f(f^-1(x)) = f((x-3)/(2)) f(f^-1(x)) = 2((x-3)/(2)) + 3 f(f^-1(x)) = (x-3) + 3 f(f^-1(x)) = x Step 2: Calculate f^-1(f(x)). f^-1(f(x)) = f^-1(2x+3) f^-1(f(x)) = ((2x+3)-3)/(2) f^-1(f(x)) = (2x)/(2) f^-1(f(x)) = x Step 3: Compare the results. Since f(f^-1(x)) = x and f^-1(f(x)) = x, it is proven that f(f^-1(x)) = f^-1(f(x)). 2.3.6 Draw a neat graph of f(x) and f^-1(x) on the same system of axes on ADDENDUM A (attached), clearly showing the nature of the start and endpoints on the graphs. Also show the line about which f(x) reflects f^-1(x) and label it. To draw the graph: • Graph of f(x) = 2x+3: • Plot the starting point: For x=-1, f(-1) = 2(-1)+3 = 1. This is a closed circle at (-1, 1) because x -1. • Plot the ending point: For x=2, f(2) = 2(2)+3 = 7. This is an open circle at (2, 7) because x < 2. • Draw a straight line connecting these two points. • Graph of f^-1(x) = (x-3)/(2): • The domain of f^-1(x) is x [1, 7). • Plot the starting point: For x=1, f^-1(1) = (1-3)/(2) = -1. This is a closed circle at (1, -1). • Plot the ending point: For x=7, f^-1(7) = (7-3)/(2) = 2. This is an open circle at (7, 2). • Draw a straight line connecting these two points. • Line of reflection: • Draw the line y=x. This line acts as the axis of symmetry between f(x) and f^-1(x). Label this line as y=x. 3 done, 2 left today. You're making progress.