Let f(x) = 2x + 5 and g(x) = 3x − 8, determine the domain of h (x) = g(x) = f(x) ○ (-∞, -)U (-7, ∞) ○ (-∞, -/-)U (⁄2, ∞) 0 (-∞, U (, ∞0) ○ (-∞, -)U ( − 1, ∞)
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- The function f(x) has a domain of (-∞, -1) u (-1, ∞) and a second derivative given by the following. -8(x + 7)¹¹(x-9)²(x + 1) (x + 1) f"(x) X = = Find the x-value(s) of the inflection points of f(x). (Enter your answers as a comma-separated list. If there are no inflection points, enter NONE.)Determine the interval(s) over which the graph of f(x) = -x6-6x5+12x-2 is concave up or concave down. Concave up on (-∞, -4)U(0, ∞) Concave down on (-4, 0) Concave up on (-4, 0) Concave down on (-∞, -4)U(0, ∞) Concave up on (0,4) Concave down on (-∞, 0)U(4, ∞) Concave up on (-∞, 0) U (4, ∞) Concave down on (0,4)Identify the equation that defines y as a function of x (a) x+(y- 1)² = 3 (b) x2-y = 3xy (c) 3x+ 5y = 1 (d) x*+y* = 1 Select one: (d) O (b) (c) O (a)
- Suppose f(z) =z'+ 3z +1. Show that f has exactly one root (or zero) in the interval [-6, -1]. Student solution. First, we show that f has a root in the interval (-6, -1). Since f is a choose function on the interval [-6, -1] and f(-6) = and f(-1) = the graph of y = f(z) must cross the z-axis at some point in the interval (-6, -1) by the choose Thus, f has at least one root in the interval [-6, -1] Second, we show that f cannot have more than one root in the interval [-6, -1]. Suppose that there were two roots z = a and z = b in the interval [-6, –1] with a < b. Then we have f(a) = f(b) = Since f is choose on the interval [-6, -1] and choose on the interval (-6, -1), by choose there would exist a point c in interval (a, b) so that f'(c) = 0. However, the only solution to f'(z) = 0 is which is not in the interval (a, b), since (a, b) C-6, -1]. Thus, f cannot have more than one root in [-6, -1]. Note Where the problem asks you to make a choice select the weakest choice that works in the…f(x) = 5x³15x² + x - 30 Select the correct statement/s: A. f(x) is strictly convex on (-1, ∞) B. C. D. When x-3, the function is increasing f(x) is strictly concave on (-∞, 1) f'(x) is not linearConsider the function f(x) = 6x + x² and the point P(-2, -8) on the graph of f. (a) Graph f and the secant lines passing through P(-2, -8) and Q(x, f(x)) for x-values of -3, - 2.5, - 1.5. 10- 10 8- 8 6- y 4 6, y 2- 24 -4 -2 0 -2 -10 -8 6 2 4 10 -10 -8 6 4 4 10 -2A -10 10 8 8- -10 -8 6 -4 -2 12 8 10 -10 -8 -6 -2 0 2 -4 4 8 10 -2 X -4 -8 -10 -10 (b) Find the slope of each secant line. -9 X (line passing through Q(-3, f(x))) 3 -8.75 (line passing through Q(-2.5, f(x))) 4 -6.75 (line passing through Q(-1.5, f(x))) (c) Use the results of part (b) to estimate the slope of the tangent line to the graph of f at P(-2, -8). 5 1
- Let f(x)= 3x(x-1)2 be a function defined on the interval [1,∞). Find the solution to f(x)= 78.Define f(x ) = Vx²+ 1 and g(x) = 2 – x². - State the domain of (f + g)(x æ # v2, -V2 [-1, 00) (-∞, 0) x # v2 (-00, -1]Assume that the function f is a one-to-one function. (a) If f(5) = 2, find f-'(2). Your answer is (b) If ƒ-( – 4) 4, find f( – 4). Your answer is If f(x) = x + 7 and g(x) = x – 7, (a) f(g(x)) = 1 (b) g(f(x)) (c) Thus g(x) is called an function of f(x) S PhotoGrid