Graph analysis of g(x)

Here are the solutions to your tutorial questions: 1) Graph analysis of g(x) a) State the values of g(-2), g(0), g(2), and g(3). From the graph: • g(-2) = 0 • g(0) = -1.5 • g(2) is undefined (open circle at (2,3)) • g(3) = 2.5 b) For what Value(s) of x is g(x) = 3? • g(x) = 3 only at x = -4 (filled circle). The function approaches y=3 at x=2 and x=4 but does not reach it due to open circles. The value is x = -4. c) For what value(s) of x is g(x) < 3? • The function is defined on [-4, 2) [2.5, 4). • At x=-4, g(-4)=3, so it's not included. • For all other points in the domain, g(x) < 3. The values are x (-4, 2) [2.5, 4). d) State the domain and range of g. • Domain: The set of all x-values for which the function is defined. Domain: [-4, 2) [2.5, 4) • Range: The set of all y-values the function takes. The lowest point is y=-1.5 and the highest value reached or approached is y=3. Range: [-1.5, 3] e) On what interval(s) is g increasing? • The function increases from approximately x=0.5 to x=2. • The function increases from x=2.5 to x=4. The intervals are (0.5, 2) and (2.5, 4). 2) Determine whether the equation defines y as a function of x. A relation defines y as a function of x if for every input x, there is exactly one output y. a) 3x - 5y = 7 Step 1: Solve for y. 5y = 3x - 7 y = (3)/(5)x - (7)/(5) Step 2: For every x, there is exactly one y. This equation defines y as a function of x. b) 3x^2 - 2y = 5 Step 1: Solve for y. 2y = 3x^2 - 5 y = (3)/(2)x^2 - (5)/(2) Step 2: For every x, there is exactly one y. This equation defines y as a function of x. c) x^2 + (y-3)^2 = 5 Step 1: This is the equation of a circle centered at (0,3) with radius sqrt(5). Step 2: For a given x (e.g., x=0), (y-3)^2 = 5 y-3 = ±sqrt(5) y = 3 ± sqrt(5). There are two y values for a single x. This equation does not define y as a function of x. d) 2x - |y| = 0 Step 1: Solve for |y|. |y| = 2x Step 2: For x > 0, y = ± 2x. For example, if x=1, y=2 or y=-2. There are two y values for a single x. This equation does not define y as a function of x. 3) Define the following. a) Function f(x) A function f(x) is a rule that assigns to each element x in a set called the domain, exactly one element, denoted f(x), in a set called the codomain. b) Domain of a function f(x) The domain of a function f(x) is the set of all possible input values (x) for which the function is defined in the real numbers. c) Range of a function f(x) The range of a function f(x) is the set of all possible output values (f(x) or y) that the function can produce for the given domain. 4) Determine whether the curve below is the graph of a function of x. If it is, state the domain and range of the function. First graph (sideways parabola): • This curve is not a function of x. • Explanation: It fails the vertical line test, meaning a vertical line can intersect the curve at more than one point. Second graph (step function): • This curve is a function of x. • Explanation: It passes the vertical line test, meaning any vertical line intersects the graph at exactly one point. • Domain: The x-values covered are from -2 to 2, inclusive. Domain: [-2, 2] • Range: The y-values taken are discrete integers. Range: \-2, -1, 0, 1, 2\ 5) Evaluate f(-3), f(0), and f(2) for the piecewise defined functions below. Then sketch the graph of the function. a) f(x) = x^2+2 & if x < 0 \\ x & if x 0 Step 1: Evaluate f(-3). Since -3 < 0, use f(x) = x^2+2. f(-3) = (-3)^2 + 2 = 9 + 2 = 11 Step 2: Evaluate f(0). Since 0 0, use f(x) = x. f(0) = 0 Step 3: Evaluate f(2). Since 2 0, use f(x) = x. f(2) = 2 Step 4: Sketch the graph. • For x < 0, the graph is a parabola y=x^2+2, opening upwards, approaching (0,2) from the left (open circle at (0,2)). • For x 0, the graph is a line y=x, starting at (0,0) (filled circle at (0,0)) and extending to the right. The graph has a jump discontinuity at x=0. b) f(x) = -1 & if x 1 \\ 7-2x & if x > 1 Step 1: Evaluate f(-3). Since -3 1, use f(x) = -1. f(-3) = -1 Step 2: Evaluate f(0). Since 0 1, use f(x) = -1. f(0) = -1 Step 3: Evaluate f(2). Since 2 > 1, use f(x) = 7-2x. f(2) = 7 - 2(2) = 7 - 4 = 3 Step 4: Sketch the graph. • For x 1, the graph is a horizontal line y=-1, ending at (1,-1) (filled circle at (1,-1)). • For x > 1, the graph is a line y=7-2x. At x=1, y=7-2(1)=5. So it starts at (1,5) (open circle at (1,5)) and extends to the right with a slope of -2. The graph has a jump discontinuity at x=1. 6) Determine whether f is even, odd, or neither. a) f(x) = (x)/(x^2+1) Step 1: Find f(-x). f(-x) = ((-x))/((-x)^2+1) = (-x)/(x^2+1) Step 2: Compare f(-x) with f(x) and -f(x). f(-x) = -((x)/(x^2+1)) = -f(x) The function is odd. b) f(x) = (x^2)/(x^4+1) Step 1: Find f(-x). f(-x) = ((-x)^2)/((-x)^4+1) = (x^2)/(x^4+1) Step 2: Compare f(-x) with f(x). f(-x) = f(x) The function is even. c) f(x) = 1 + 3x^3 - x^5 Step 1: Find f(-x). f(-x) = 1 + 3(-x)^3 - (-x)^5 = 1 - 3x^3 + x^5 Step 2: Compare f(-x) with f(x) and -f(x). • f(-x) f(x) (e.g., 1 - 3x^3 + x^5 1 + 3x^3 - x^5) • f(-x) -f(x) (e.g., 1 - 3x^3 + x^5 -(1 + 3x^3 - x^5) = -1 - 3x^3 + x^5) The function is neither even nor odd. 7) Find the functions (i) f g, (ii) g f, (iii) f f, (iv) g g. a) f(x) = x^3 + 5, g(x) = [3]x (i) f g (x) = f(g(x)) = f([3]x) = ([3]x)^3 + 5 = x + 5 (ii) g f (x) = g(f(x)) = g(x^3 + 5) = [3]x^3 + 5 (iii) f f (x) = f(f(x)) = f(x^3 + 5) = (x^3 + 5)^3 + 5 = x^9 + 15x^6 + 75x^3 + 130 (iv) g g (x) = g(g(x)) = g([3]x) = [3][3]x = [9]x b) f(x) = (1)/(x), g(x) = 2x + 1 (i) f g (x) = f(g(x)) = f(2x+1) = (1)/(2x+1) (ii) g f (x) = g(f(x)) = g((1)/(x)) = 2((1)/(x)) + 1 = (2)/(x) + 1 = (2+x)/(x) (iii) f f (x) = f(f(x)) = f((1)/(x)) = (1)/(1)x = x (iv) g g (x) = g(g(x)) = g(2x+1) = 2(2x+1) + 1 = 4x + 2 + 1 = 4x + 3 8) Find f g h. a) f(x) = 3x - 2, g(x) = x, h(x) = x^2 Step 1: Find g(h(x)). g(h(x)) = g(x^2) = (x^2) Step 2: Find f(g(h(x))). f(g(h(x))) = f((x^2)) = 3(x^2) - 2 The composite function is 3(x^2) - 2. b) f(x) = x, g(x) = (x)/(x-1), h(x) = [3]x Step 1: Find g(h(x)). g(h(x)) = g([3]x) = [3]x[3]x-1 Step 2: Find f(g(h(x))). f(g(h(x))) = f([3]x[3]x-1) = ([3]x[3]x-1) The composite function is ([3]x[3]x-1). 9) Use the table below to evaluate each expression. | x | 1 | 2 | 3 | 4 | 5 | 6 | |---|---|---|---|---|---|---| | f(x) | 3 | 1 | 5 | 6 | 2 | 4 | | g(x) | 5 | 3 | 4 | 1 | 3 | 2 | a) f(g(3)) Step 1: Find g(3) from the table. g(3) = 4. Step 2: Find f(4) from the table. f(4) = 6. f(g(3)) = 6 b) g(f(2)) Step 1: Find f(2) from the table. f(2) = 1. Step 2: Find g(1) from the table. g(1) = 5. g(f(2)) = 5 c) (f g)(5) Step 1: Find g(5) from the table. g(5) = 3. Step 2: Find f(3) from the table. f(3) = 5. (f g)(5) = 5 d) (g f)(5) Step 1: Find f(5) from the table. f(5) = 2. Step 2: Find g(2) from the table. g(2) = 3. (g f)(5) = 3 e) g(g(g(2))) Step 1: Find g(2) from the table. g(2) = 3. Step 2: Find g(3) from the table. g(3) = 4. Step 3: Find g(4) from the table. g(4) = 1. g(g(g(2))) = 1 f) (f f f)(1) Step 1: Find f(1) from the table. f(1) = 3. Step 2: Find f(3) from the table. f(3) = 5. Step 3: Find f(5) from the table. f(5) = 2. (f f f)(1) = 2 10) Assume that f is a one-to-one function. For a one-to-one function, if f(a)=b, then f^-1(b)=a. a) If f(6) = 17, what is f^-1(17)? Using the property of inverse functions, if f(6)=17, then f^-1(17) is the input that gives 17 as output, which is 6. f^-1(17) = 6 b) If f^-1(2) = 2, what is f(2)? Using the property of inverse functions, if f^-1(2)=2, then f(2) is the output when the input is 2, which is 2. f(2) = 2 3 done, 2 left today. You're making progress.