Question

What is the equation of the tangent line of `f(x)=sqrt(x-3)/x^4` at `x=3`?

Answer 1

`x > 3`, to make f real. So, x = 3 is not a point for tangency at all.

Explanation:

The function is not continuous at x = 3, and so, not differentiable at

all.

Interestingly, the graph has distinctive features

x>0

x-axis `rarr` is the asymptote

The turning point ( f' = 0 ) is at (24/7, 0.005)#, nearly

Max f = 0.005, nearly.

With aptly chosen units, the Socratic graph reveals all these

features.

graph{sqrt(x-3)/x^4 [-25.35, 25.34, -.01, .01]}

Answer 2

The tangent line is the vertical line `x=3`

Explanation:

`f'(x) = (18-5x)/(2x^4sqrt(x-3))`

`lim_(xrarr3^+)abs(f(x)) = oo`, so the tangent line at `x=3` is vertical.

The vertical line at `x=3` has equation `x=3`.