Saved Bookmarks
| 1. |
Find the points of local maxima or local minima, if any, of the functions, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be: f(x) = x3 – 6x2 + 9x +15 |
|
Answer» Given, as f(x) = x3 – 6x2 + 9x + 15 On differentiating with respect to x, we get, f‘(x) = 3x2 – 12x + 9 = 3(x2 – 4x + 3) = 3 (x – 3) (x – 1) For all the maxima and minima, f’(x) = 0 = 3(x – 3) (x – 1) = 0 = x = 3, 1 At x = 1 f ’(x) changes from negative to positive Since, x = – 1 is a point of Maxima At x =3 f‘(x) changes from negative to positive Since, x =3 is point of Minima. Thus, local maxima value f (1) = (1)3 – 6(1)2 + 9(1) + 15 = 19 Local minima value f (3) = (3)3 – 6(3)2 + 9(3) + 15 = 15 |
|