(a) Find the volume of the solid generated by revolving the region bounded by the curves
y = x² and y = 4x - x² about the line y = 6
(b) Sketch the graph of a continuous function f(x) satisfying the following properties:
(i) the graph of f goes through the origin
(ii) f¹(-2) = 0 and f¹(3) = 0.
(iii) f¹(x) > 0 on the intervals (-∞, -2) and (-2; 3).
(iv) f¹(x) < 0 on the interval (3;∞).
Label all important points.
1
Expert's answer
2020-07-02T19:17:14-0400
a)
y=x2,y=4x−x2,y=6∴x2=4x−x2→x2−2x=0→x=0,2∴V=π∫02r12−r22dxr1=6−x2,r2=6−4x+x2∴V=π∫02(6−x2)2−(6−4x+x2)2dx=π∫0236−12x2+x4−(x4−8x3+28x2−48x+36)dx=π∫0236−12x2+x4−x4+8x3−28x2+48x−36dx=π∫028x3−40x2+48xdx=π(2x4−340x3+24x2)∣∣02∴V≈67b)f′(x)>0in(−∞,−2),(−2,3)therefore the function increasingf′(x)<0in(3,∞)therefore the function decreasingf′(x)=0atx=−2,x=3Therefore the tangent at these points is horizontal
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