AB challenging the calculation problems? The best response?
The graph of the function f is composed of three line segments with points (-2.0) (1.4), (2.1), and (4, -1). It gives an image that looks like a kind of top of a triangle (the increase (-2.0) to (1.4) and decreased of (1.4) to (2.1) and decreased again (2.1) to (4, -1). A) Let g (x) = integral with upper and lower limit x 1: f (t) dt a Calculate g) (4) g (-2). b) Calculate the instantaneous rate of change of g with respect to x, at x = 1. c) Determine the absolute minimum value of g on the closed interval [-2.4]. Justified your answer. D) The second derivative of g is not defined at x = 1 and x = 2. How many of these values are the coordinates of inflection or points on the graph of g? response. justify answers: a) g (4) = 2.5 g (-2) =- 6 b) 4 c) The minimum value is -6. D) Can you explain how to do these problems? Thanks!
I’ll have a go but I’m sort of work these as I go. You can work the equations for the three line segments and then to each of integrals according to x values for which they have to solve g. Or … Draw the graph. If you are working in the area under the curve, you can see it is mainly composed of triangles. The area of a triangle is half the size x times y-dimension. Of -2 to 1 The area is: (3 x 2.4) = 6 1 to 2 the area is: (3 x 1 / 2) + 1 = 5.2 2 to 4, the area is: 0 (two opposite triangles) a), g (4) = Area 1 to 4 = 2.5 (top) g (-2) = negative (-2 to 1 area ) = -6 (from above) b) The exchange rate is the derivative instant. The derivative of an integral is the original function. Therefore, g ‘(1) = f (1) = 4 c) OK Now I see when we left, the integral is negative. Therefore, g decreases steadily as we move towards the left, and reaches a minimum where f touches the x axis at x =- 2. g (-2) = -6. Not sure if this would be a fairly formal response. I guess you can see that g is positive for x> 1. d) The need to think carefully on the definition of a turning point! One definition is that is when the first derivative is maximum or minimum. We have seen that the first derivative is the function original f This in a maximum at x = 1, but not at x = 2. Therefore, there is only one turning point at x = 1.
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