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1.4 f(t) = sin((1+0.5)2) {osts 2}| 2 PARTA 1-2 t = 01 If we define a new function g as follows then g(x) is the area under the graph off from 0 to x [until f(x) becomes negative, at which point g(x) becomes a difference of areas]. Use the b slider to determine the value of x at which g(x) starts to decrease. [Unlike the integral in Question 2, it is impossible to evaluate the integral defining g to obtain an explicit expression for g(x).] t = 1.27 g(x) = 0.85333978 0.8 f(t)dt -0.6 x b= 1.27 -0.4 2 PART B 7 -0.2 Use the b slider to sketch the region under the graph off from 0 to the values of x in the table below, recording the corresponding values of g(x) as you go -0.2 0 02 04 06 08 12 14 16 18 * Nyered 0 0 -0.21 0.2 .07090041 0.4 19050599 0.0 .35789409 -0.4 0.8 .55457174 1 .73075076 -0.6 1.2 .84422121 1.4 .62455555 1.0 07058553 -0.8 1.8 48405278 2 38903 -1 Click on the circle above the table and next to g(x) to shock the accuracy of your sketch -1.2 PARTC Use the graph of g from PART B to fill in the table below, using the interpretation of g'(x) as the slope of the tangent line. (You can click on the circles and check the label box below for visualization.) y=g'(b)(x - 6)+(6)0 - 2