1. tangent line: (y - ) = (x - ) e2
f(x) =
ln x x
f’(x) =
1 - ln x x2
Part (a)
mtan =
1 - ln e2
(e2)2
1 - 2 e4 = -1 e4 =
f(e2) =
ln e2 e2 2 e2 =
tangent line: (y ) = (x - ) 2 e2 -1 e4 e2

2. Part (b) = 0 1 – ln x = 0 ln x = 1 x = e ln x 1 - ln x f’(x) = f(x) =
Multiply both sides by x2.
1 - ln x x2
= 0
1 – ln x = 0
ln x = 1
First, test a value less than e. x=1 is a convenient choice.
x = e
f’(1) =
1 - ln 1 12
= 1 e
Since f’(x) is positive, the graph of f(x) is increasing on this interval.

3. Part (b) f(x) = ln x x f’(x) = 1 - ln x x2
Next, test a value greater than e. x = e2, x = e3, etc. would be convenient choices.
f’(e2) =
1 - ln e2
(e2)2 = e4
1 - 2 = e4 -1
Since f’(x) is negative, the graph of f(x) is decreasing on this interval. e
There is a RELATIVE MAX at x = e, since f(x) goes from increasing to decreasing there.

4. Use the Product Rule to find the 2nd derivative.
f(x) =
ln x x
f’(x) =
1 - ln x x2
Part (c) v u
f’(x) =
1 - ln x x2
Use the Product Rule to find the 2nd derivative.
f”(x) =
(-1/x)(x2) – (1 - ln x)(2x) x4
Set f”(x) = 0.
x = 0 or 2 ln x = 3
-x – 2x + 2x ln x = 0
(not in the domain)
-3x + 2x ln x = 0
ln x = 3/2
x (2 ln x – 3) = 0
x = e3/2

5. Part (d) lim - or DNE ln x 1 - ln x f’(x) = f(x) = x x2
(approaching -)
(positive,approaching 0)
f(x) =
ln (0.0001)
0.0001
lim
x0+
- or DNE