Example 1 (Second derivative test)Let f(x) = (x^2  3x  3)e^x. Then f'(x) = (x^2  x  6)e^x,and f''(x) = (x^2 + x  7)e^x. The critical points are at x= 2 and x = 3. Note that f''(2) = 5 e^{2} < 0 and f''(3) = 5e^3 > 0. So f has a local max at x =2 and a local min at x = 3. Example 1 (Second derivative test)Let f(x) = (x^2  3x  3)e^x. Then f'(x) = (x^2  x  6)e^x,and f''(x) = (x^2 + x  7)e^x. The critical points are at x= 2 and x = 3. Note that f''(2) = 5 e^{2} < 0 and f''(3) = 5e^3 > 0. So f has a local max at x =2 and a local min at x = 3. 
Example 2 (Second derivative test)Let f(x) = 3x^5 + 5x^4. Then f'(x) = 15x^4 + 20x^3 = 5x^3(3x+ 4),and f''(x) = 60x^3 + 60 x^2. The critical points are at x= 0 and x = 4/3. Note that f''(4/3) = 320/9 < 0 and f''(0) = 0. So f has a local max at x = 4/3. The second derivative test doesn't say anything about what's happening at x = 0. Example 2 (Second derivative test)Let f(x) = 3x^5 + 5x^4. Then f'(x) = 15x^4 + 20x^3 = 5x^3(3x+ 4),and f''(x) = 60x^3 + 60 x^2. The critical points are at x= 0 and x = 4/3. Note that f''(4/3) = 320/9 < 0 and f''(0) = 0. So f has a local max at x = 4/3. The second derivative test doesn't say anything about what's happening at x = 0. 
Example 3 (Curve sketching)f(x) = 3x^3 + 3x^2  4x + 1.Example 3 (Curve sketching)f(x) = 3x^3 + 3x^2  4x + 1. 
Example 4 (Curve sketching)f(x) = \frac{3x  1}{4x + 1}.
Example 4 (Curve sketching)f(x) = \frac{3x  1}{4x + 1}.

Example 5 (Curve sketching)f(x) = \frac{3x  1}{x^2+ 1}.
Example 5 (Curve sketching)f(x) = \frac{3x  1}{x^2+ 1}.

