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- Prove, without graphing, that the graph of y= x^2 − 3 + 1/x. ? has at least two x-intercepts in the interval (0,2).3. Find the slope of the graph tangent to the curve y = -x* – 2x3 +x that passes 6. through the point (2, -9)The point P(5,-2) lies on the curve y = (a) If Q is the point (x, (1) 4.9 MpQ (ii) 4.99 (iii) 4.999 mpQ = (iv) 4.9999 MpQ (v) 5.1 MpQ (vi) 5.01 MpQ (vii) 5.001 mpQ nt (x₁__²_x). 2 m = 2 4-x find the slope of the secant line PQ (correct to six decimal places) for the following values of x. (viii) 5.0001 = MpQ² (b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at P(5,-2). (c) Using the slope from part (b), find an equation of the tangent line to the curve at P(5,-2).
- 37. Find the coefficients a, b, c, and d so that the curve shown in the accompanying figure is the graph of the equation y = ax' + bx? +cx +d. 20 (0, 10) (1, 7) 6. (3, -11) (4,-14) -20 Figure Ex-37The line whose slope is 3 is tangent to the curve y=x² + 3x + 2 at the point O (3,-2) O (-3, 2) O (3, 2) O (-3, -2)find the slope of the curve at the given points (x2 + y2)2 = (x - y)2 at (1, 0) and (1, -1)
- To find the length of the curve defined by y = 4x + 10x from the point (-3,2886) to the point (4,16424), you'd have to compute * where a 0b9.0319-1058, and f(x) = N|M a f(x)dx 416424 + -416424¹¹ +48-416424³ +100-416424 576 11Suppose 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…To find the length of the curve defined by from the point (-2,372) to the point (1,12), you'd have to compute where a b and f(x) y = 6x6 + 6x - •b a f(x) dx