if the gradient of a curve y = x^3 at a point is 12 find the points?

Answer 1

dy/dx= 3x^2=12
therefore x=2 or -2
and y=8 or -8

Answer 2

f(x) = 6x^2 - x^3 f'(x) = 12x - 3x^2 12 = 12x - 3x^2 4 = 4x - x^2 x^2 - 4x + 4 = 0 (x - 2) * (x - 2) = 0 x = 2 a) x = 2 b) y = 6 * 2^2 - 2^3 = 6 * 4 - 8 = sixteen y = mx + b m = 12 x = 2 y = sixteen sixteen = 12 * 2 + b sixteen = 24 + b -8 = b y = 12x - 8

Answer 3

Gradients are vectors not scalars. You're question doesn't make sense.

Answer 4

gradient is (3x^2,-1); using x=12 gives (432,-1)

or (-3x^2,1) and (-432,1)