posted by jane .
Let f(x) = 2/3-x
a) find the slope of the tangent to the graph of f at a general point x0 using the definition of the dervative.
b) use the result in part a to find the slope of the tangent at x0=1.
I am very confused.
My answer for (a) 1/ 6-6x-x^2
I compared them to a classmate - answers didn't agree.
I am lost.
How did you get that derivative?
I rewrote the equation to
y = 2(3-x)^-1
then y' = -2(3-x)^-2(-1)
= 2/(3-x)^2 or 2/(x^2 - 6x + 9)
so when x=1, y' = 2/(3-1)^2 = 1/2