crazco said:Homework Statement
At x=3 does the function y=x^3 + (x-3)^(1/3) have a vertical tangent, cusp, corner or none?
The Attempt at a Solution
I took the derivative
y'=3x^2 + 1 / 3x^(2/3)
then i replaced 3 in which gave 27 + 1 / 0
I don't understand how to come to the correct conclusion following this step.
Vertical tangent comes to mind since 1 / 0 is a vertical line, but I don't know how to prove it using limits.
A cusp in a function occurs at a point where the function has a sharp turn and the derivative is undefined. This results in a point where the graph of the function has a "pointy" appearance.
To determine if a function has a cusp at a point, you can look at the behavior of the function and its derivative at that point. If the function has a sharp turn and the derivative is undefined at that point, then it has a cusp.
A vertical tangent in a function occurs at a point where the slope of the function is infinite, resulting in a vertical line. This usually happens at a point where the derivative is zero.
To determine if a function has a vertical tangent at a point, you can look at the slope of the function and its derivative at that point. If the slope is infinite and the derivative is zero, then it has a vertical tangent.
A corner in a function occurs at a point where the function has a sharp turn and the derivative is defined. This results in a point where the graph of the function has an abrupt change in direction.
To determine if a function has a corner at a point, you can look at the behavior of the function and its derivative at that point. If the function has a sharp turn and the derivative is defined at that point, then it has a corner.