The function f(x)=(x-3)^(1/3) has a vertical tangent line...?

Question

The function f(x)=(x-3)^(1/3) has a vertical tangent line...

find the point where the vertical tangent line occurs algebraically

Answer 1

Take the derivative and see where it is undefined. That will be where you wind up dividing by zero.

f(x) = (x - 3)^(1/3)

f'(x) = (1/3)(x - 3)^(-2/3) = 1/[3(x - 3)^(-2/3)]

The denominator equals zero at x = 3.

The vertical tangent is at the point (x, f(x)) = (3, 0).

Answer 2

the function has a vertical tngt if the first derivate of the function is ±∞

The first derivate is: [(x-3)^(1/3)] / [3x-9]
which will be =+∞ if the denominator =0

So the denominator will be =0:
3x-9=0
3x=9
x=9/3=3

so: at x=3 f(x) has a vertical tangent line