Page 57 - Calc for the Clueless
P. 57
No third derivative.., too messy!
2
Intercept (0,0). No vertical asymptote since x + 1 = 0 has only imaginary roots. Horizontal asymptote y = 0
since the degree of the top is less than the degree of the bottom.
2
Possible max, min: y' = 0 1 - x = 0, x = ±1. Substituting in the original, we get (1,2) and (-1,-2). f"(1) is
negative. (1,2) is a maximum, f"(-1) is positive. (-1,-2) is a minimum.
2
Possible inflection points: y" = 0, -8x(3 - x ) = 0. x = 0, . Substituting into the original, we get the points
(0,0), .
By the messier test, all three are inflection points.
If you liked that one, you'll love this one.
Example 30—
There are no asymptotes. Intercepts: y = 0 (0,0), (4,0). y' = 0, x = 1,4. Substituting in the original, we get (1,9),
(4,0). f"(1) is negative. (1,9) is a maximum, f"(4) is positive. (4,0) is a minimum.
