Asked by Anonymous
Determine the x coordinate of the point for which the tangent is horizontal to the function f(x) = (2^x)/x
Answers
oobleck
using the quotient rule,
f'(x) = 2^x (x ln2 - 1)/x^2
so f'(x) = 0 at x = 1/ln2
at that point,
f(1/log2) = e ln2
(because 1/ln2 = log<sub><sub>2</sub></sub>e)
f'(x) = 2^x (x ln2 - 1)/x^2
so f'(x) = 0 at x = 1/ln2
at that point,
f(1/log2) = e ln2
(because 1/ln2 = log<sub><sub>2</sub></sub>e)