Thread: Implicit Differentiation & Horizontal tangents

Given the function find the derivative and horizontal tangent(s).


I've obtained the derivative as


However i am confused about the horizontal tangent, i understand that a tangent is horizontal when , but i am unsure of what to do next.



I'm not after the answer, but the next step(s) required.

It's tough to supply a next step when you already have.

Find where the numerator is zero and the denominator is not.

You may wish to try it after completing the square.



Maybe not. It was just a thought.

Anyway, since there is nowhere but the Origin where both numerator and denominator are zero, just solve the numerator. Obviously, you will get a line, not a point. Substitute into the original equation and get the two points.

Originally Posted by JohnnyB

Given the function find the derivative and horizontal tangent(s).


I've obtained the derivative as


However i am confused about the horizontal tangent, i understand that a tangent is horizontal when , but i am unsure of what to do next.



I'm not after the answer, but the next step(s) required.

if , then ...

substitute for in the equation for the curve and solve for ... then determine the corresponding y-value for each x solution you find.

Probably a stupid question...

but why is the numerator of the derivative used and not the denominator, or both?

Originally Posted by JohnnyB

Probably a stupid question...

but why is the numerator of the derivative used and not the denominator, or both?

if the denominator is zero it is undefined
if the numerator is zero you have zero for an answer