f'(x) = 1/(2x3 + 5) rewrite f'(x) = (2X3 + 5) -1 use the chain rule d/dx (2x3 + 5) - 1 -1 * (2x3 + 5)-2 * 6x2 - 6x2(2x3 + 5) -2 ==================I would leave like this rather than rewriting this
2x(x + 5)(x - 2)
Since there is nothing following, none of them!
2 x 3 = 65 x -5 = -256 + -25 = -19
Rearrange: 4x5 + 6x2 + 6x3 + 9 Group: 2x2 (2x3 + 3) + 3 (2x3 + 3) Simplify to get your answer: (2x2 + 3) (2x3 + 3)
f'(x) = 1/(2x3 + 5) rewrite f'(x) = (2X3 + 5) -1 use the chain rule d/dx (2x3 + 5) - 1 -1 * (2x3 + 5)-2 * 6x2 - 6x2(2x3 + 5) -2 ==================I would leave like this rather than rewriting this
your equation is this... 2x3 + 11x = 6x 2x3 + 5x = 0 x(2x2 + 5) = 0 x = 0 and (5/2)i and -(5/2)i
60
(x + 1) / (2x3 + 3x + 5) --- Can't depict long division here, but you can work this out by seeing if (x + 1) is a factor of (2x3 + 3x + 5). If it is, then the answer will be 1/(the other factor). In this case, that is true. (2x3 + 3x + 5)/(x + 1) = 2x2 - 2x + 5, so the term term above is equal to: 1 / (2x2 - 2x + 5)
2x(x + 5)(x - 2)
Since there is nothing following, none of them!
2 x 3 = 65 x -5 = -256 + -25 = -19
Rearrange: 4x5 + 6x2 + 6x3 + 9 Group: 2x2 (2x3 + 3) + 3 (2x3 + 3) Simplify to get your answer: (2x2 + 3) (2x3 + 3)
13
(x + 5)2
2x3 - 7 + 5x - x3 + 3x - x3 = 8x - 7
6