Question 1: 1. The problem statement, all variables and given/known data Find "b" so that the difference between the x co-ordinates of the two inflection points of y=x4 + 4x3+bx2+5x+7 is 3 3. The attempt at a solution I.P. at y' = 0 two x co-ordinates are x and x+3 y' = 4x3 + 12x2 + 2bx +5 0 = 4x3 + 12x2 + 2bx +5 0 =4(x+3)3 + 12(x+3)2 + 2b(x+3) +5 4x3 + 12x2 + 2bx +5 = 4(x+3)3 + 12(x+3)2 + 2b(x+3) +5 expand, add, subtract and solve for x, you get: (-216-6b)/72 = x I.P is y''=0 y'' = 12x2 + 24x +2b sub in x 0= 12[(-216-6b)/72]2+24 (-216-6b)/3]+2b then you expand, add, subtract and end up with the quadratic equation 0= 36 (b2+48b +432) use the quadratic formula to get b=-12 or or b=-36 to check if b is right, I plugged it back in to y'' 0= 12x2 + 24x -2b 0=12x2 + 24x -2(-12) 0 = 12x2 + 24x -24 x=-3.8284 or x=1.8284 but the difference here is 5.6568 0=12x2 + 24x -2(-36) x= -1, x = 3 the difference here is 4. Question 2: 1. The problem statement, all variables and given/known data if f(x) = x3 + 3x2 + k has three distinct real roots, what are the bounds on "k"? (i.e., ? <x<?). Hint: Look for extema using f' and f''. 3. The attempt at a solution f(x) = x3 + 3x2 + k f'(x)= 3x2 +6x 0= 3x(x+2) x = 0, x=-2 f"(x) = 6x+6 f'' (0) = 6 at x=0, the function is concave up f''(-2) = -6 at x = -2, it is concave down because of the type of graph (its a cubic function), the y-int (which is k) must be equal to or less than 0, but has to be greater than equal to -4... as I was typing this, I realized that this doesnt make sense, so I dont understand this question. I will appreciate any help I can get.