Find the gradient of the curve y=2x^3 - 5x^2 + 46x + 87 at the point where it crosses the x axis.
I'm having real problems with this one as all the examples i've done before have given me a full set of co-ordinates.
I know that it will cross the x axis when y equals 0 so i have at least one co-ordinate. I differentiated it to 6x^2 - 10x + 46 I then thought about making 6x^2 - 10x = -46 and trying to get a value of x from that but i'm getting nowhere.
Any help on this is apprecciated
ally79, I wrote this post, if you have not yet realized that you need to do.
First, remember the definition of a gradient of the function :
The gradient of the function is a vector, whose projection on the axises of coordinates are partial derivatives of functions , i.e.
You have to find of at the point where it crosses the x axis, i.e. you need to find the absolute value of the derivative function at the point where it crosses the x axis.
Finally, you need to find this value:
using the rational root theorem and the intermediate value theorem, you can determine that there exists one real root at .Find the gradient of the curve y=2x^3 - 5x^2 + 46x + 87 at the point where it crosses the x axis.
(why teachers still make students do this is beyond me)
now ... evaluate at