Question

DETERMINE THE EQUATION OF THE NORMAL LINE TO THE CURVE Y=2^(1/X) AT THE POINT WHERE X=1.

Answer 1

let's find our point first
when x = 1, y = 2^1 = 2
so we have the point (1,2).

The derivative gives us the slope of the tangent.
y' = ln2(2^(1/x))(-1/x^2)
when x = 1
y' is ln2(2)(-1) = -2ln2

so the slope of the normal is +1/(2ln2)

so the equation of that normal is
x - 2ln2 y = c
at (1,2)
1 - 2ln2(2) = c
c = 1 - 4ln2

so the normal equation is
x - 2ln2 y = 1 - 4ln2