Question

(x+6)³+(3x+2)³\(x+6)³-(3x+2)³​

Answer 1

Answer:

f '(x) = 2(3x2 + 5) + 6x(2x - 1)= 6x2 + 10 + 12x2 - 6x= 18x2 - 6x + 102)r(x) = 6x - 11s(x) = 8x + 1    r'(x) = 6  s'(x) = 83)                                                                                 Quotient Ruler(x) = (5x - 2)(2x + 3)s(x) = x - 4Use the product rule table 

below to get: 

      r'(x) = 20x + 11

 s'(x) = 1Product Ruler(x) = 5x - 2s(x) = 2x + 3   r'(x) = 5   s'(x) = 2f'(x) = (x - 4)(20x + 11) - (5x - 2)(2x + 3)/(x - 4)2= (20x2 +11x - 80x -44 - 10x2 -15x + 4x + 6)/(x - 4)2= (10x2 -80x - 38)/(x - 4)2= 2(5x2 - 40x - 19)/(x - 4)24)v(x) = 8x + 3f(v) = v2   v'(x) = 8  f'(v) = 2v 

or 2(8x + 3)

f '(x) = 8(2)(8x + 3)= 16(8x + 3)5)v(x) = 8 - 5xf(v) = v1/2  v'(x) = -5f '(x) = v-1/2/2 

or (8 - 5x)-1/2/2

f '(x) = -5(8 - 5x)-1/2/2= -5/2(8 - 5x)1/26)                                                                                Product ruler(x) = -6xs(x) = (7x - 1)2 r'(x) = -6s'(x) = 14(7x - 1)f '(x) = -84x(7x - 1) - 6(7x - 1)2= -6(7x - 1)(14x + 7x - 1)= -6(7x - 1)(21x - 1)7)                                                                                   Product Ruler(x) = -6s(x) = ex+1 r'(x) = 0s'(x) = ex+1 f'(x) = -6ex+1 + 0 = -6ex+18)                                                                            Natural Log Ruleg(x) = 5x2 - 3xg'(x) = 10x - 3f '(x) = (10x - 3)/(5x2 - 3x)9)                                                                                 Quotient Ruler(x) = 5x2 s(x) = ex r'(x) = 10xs'(x) = ex 13)14)To find the intersection points set the functions equal and solvex2 = xx2 - x = 0                                                        x(x - 1) = 0 means they intersect at x = 0, 1You must integrate it as the upper limit minus the lower limit.  The line is the upper limit and the parabola is the lower limit.  The area is in yellow.15)To find the intersection points, set the two functions equal and solvex3 - 3x = xx3 - 4x = 0x(x2 - 4) = 0                                              x(x - 2)(x + 2) = 0 so they intersect at x = 0, 2, -2Notice the area is divided into two parts.  Integrate them separately for two different integrals.  The left area has boundaries x = -2, x = 0 with upper boundary the curve and lower boundary the line.  The area to the right has boundaries x = 0, x = 2 with upper boundary the line and lower boundary the curve.Because this graph is symmetric, we could have integrated one of the areas and doubled the result.  We must be careful with this, however, to make sure the areas are the same size!Well that's it.  Hope you are ready for your final!!  Have a great summer!!