1. Part (a) Tangent line:(y – ) = (x – ) f’(e) = e 2 ln e m tan = e 2 (1) e2e2 2 e

2. Part (b) f’(x) = x 2 ln x Use the Product Rule to find the 2 nd derivative. uv f”(x) = 2x ln x + (x 2 )(1/x) f”(x) = 2x ln x + x f”(x) = x (2 ln x + 1) This will remain positive on the interval (1,3). Therefore, the graph is CONCAVE UP on (1,3).

3. Part (c) x 2 ln x dxf(x) = We’ll need to use integration by parts. du = 1/x dx v = 1/3 x 3 u = ln x dv = x 2 dx f(x) = 1/3 x 3 ln x- 1/3 x 2 dx f(x) = 1/3 x 3 ln x- 1/9 x 3 + C Recall from Part (a) that f(x) passes through (e,2). 2 =1/3 e 3 ln e- 1/9 e 3 + C 1 C = 2 – 2/9 e 3 2 – 2/9 e 3