Geometrical applications of calculus (1 Viewer)

[Florencee]

HELP!!! Find any stationary points on the curve y=(x-2)^4
I've already differentiated so: y'= 4(1)(x-2)^3 then what???

 

[Andy005]

To find the stationary points of a curve the gradient must be equal to zero so sub 0 into y', i.e. 4(x-2)^3 =0, then sub each of the x values into the original equation to find the y coordinates.

 

[Biblidography]

Remember to check the nature! (Sub into table of values and see whether its a min/max)

 

[pikachu975]

Remember to check the nature! (Sub into table of values and see whether its a min/max)

Don't think you need to since it just says to find the points