Question

How do you find the equation of a line tangent to the function `y=1+x^(2/3)` at (0,1)?

Answer 1

`x=0`

Explanation:

The general slope of a line is given by it's derivative.

Since `y=1+x^(2/3)`
using the exponent rule for derivatives
`color(white)("XXX")m=(dy)/(dx)=2/3 * x^(2/3-1) = 2/3 * x^(-1/3) =2/(3root(3)(x))`

Note that this slope is "undefined" (it represents a vertical line) when `x=0`
therefore the tangent line at `(x,y)=(0,1)` is
`color(white)("XXX")x=0`
(the vertical line through the point `(0,1)`)