Mathematique

Question

Let

f(x) = x^3 - 6x^2 + 9x + 1

.

f (x)

x = 0

x = 3

Answer

critical points first find the derivative:

f(x) = x^3 - 6x^2 + 9x + 1

factor it:

f^{'} (x) = 3 (x^{2} - 4 x + 3) = 3 (x - 1) (x - 3)

critical points occur where

f^{'} (x) = 0

or where the derivative is undefined. since this derivative is defined everywhere, set it equal to zero:

3 (x - 1) (x - 3) = 0

so the critical numbers are:

x = 1 and x = 3

now find the function values:

f (1) = 1 - 6 + 9 + 1 = 5

f (3) = 27 - 54 + 27 + 1 = 1

so the critical points are: p]:inline" data-streamdown="list-item">

(1, 5)

p]:inline" data-streamdown="list-item">

(3, 1)

increasing and decreasing intervals use the sign of

f^{'} (x) = 3 (x - 1) (x - 3)

. test intervals: p]:inline" data-streamdown="list-item">for

x < 1

, take

x = 0

:

f^{'} (0) = 9 > 0

so the function is increasing on

(- \infty, 1)

p]:inline" data-streamdown="list-item">for

, take x = 2 : f^{'} (2) = 3 (1) (- 1) = - 3 < 0 so the function is decreasing on (1, 3) p]:inline" data-streamdown="list-item">for x > 3, take x = 4 : f^{'} (4) = 3 (3) (1) = 9 > 0 so the function is increasing on (3, \infty) therefore: p]:inline" data-streamdown="list-item"> increasing: (- \infty, 1) \cup (3, \infty) p]:inline" data-streamdown="list-item"> decreasing: (1, 3) local maximum and minimum p]:inline" data-streamdown="list-item">at x = 1, the derivative changes from positive to negative, so there is a local maximum .p]:inline" data-streamdown="list-item">at x = 3, the derivative changes from negative to positive, so there is a local minimum . thus: p]:inline" data-streamdown="list-item"> local maximum: (1, 5) p]:inline" data-streamdown="list-item"> local minimum: (3, 1) definite integral from x = 0 to x = 3 compute: \int_{03} (x^{3} - 6 x^{2} + 9 x + 1) d x antiderivative: \int (x^{3} - 6 x^{2} + 9 x + 1) d x = \frac{x ^{4}}{4} - 2 x^{3} + \frac{9 x ^{2}}{2} + x now evaluate from 0 to 3 : [\frac{x ^{4}}{4} - 2 x^{3} + \frac{9 x ^{2}}{2} + x]_{03} at x = 3 : \frac{3 ^{4}}{4} - 2 (3^{3}) + \frac{9 ( 3 ^{2} )}{2} + 3 = \frac{81}{4} - 54 + \frac{81}{2} + 3 put everything over 4: \frac{81}{4} - \frac{216}{4} + \frac{162}{4} + \frac{12}{4} = \frac{39}{4} at x = 0 : 0 so: \int_{03} f (x) d x = \frac{39}{4}