Let f(x)=(2/3)x³-5x²+12x-7
Find f'(x) and f"(x).
f'(x) = 2x²-10x+12
f"(x) = 4x-10
Find the intervals on which f(x) is increasing or decreasing.
Set the first derivative = 0
f'(x) = 2x²-10x+12 = 0
x²-5x+6 = 0
(x-3)(x-2) = 0
x-3=0; x-2=0
x=3 x=2
This divides the number line into three intervals
------o---o-------
2 3
Intervals: | (-infinity,2) | (2,3) | (3,infinity)
Test value:| 0 | 2.5 | 4
Value of f'| 12 | -.5 | 4
Sign of f' | + | - | +
Conclusion | increasing |Decreasing| Increasing
Find the local maximum and minimum of f(x), if any
Since f(x) changes from increasing to decreasing at
x=2, and f(2) is defined, therefore there is a "peak",
or local maximum where x=2.
Since f(2) = 7/3, the local maximum point is (2,7/3)
Since f(x) changes from decreasing to increasing at
x=3, and f(3) is defined, therefore there is a "valley",
or local minimum where x=3.
Since f(3) = 2, the local minimum point is (3,2)
Find the intervals on which the graph of f (x) is concave up or concave down.
Set the second derivative = 0
f"(x) = 4x-10 = 0
4x = 10
x = 10/4
x = 5/2
This divides the number line into two intervals
-------o-------
5/2
Intervals: |(-infinity,5/2)|(5/2,infinity)
Test value:| 0 | 3 |
Value of f"| -10 | 7.5 |
Sign of f' | - | + |
Conclusion |concave down |concave up |
Find the inflection points of f(x), if any.
Since f(x) changes from concave downward to concave upward
at x=5/2, and f(5/2) is defined, therefore there is an
inflection point, where x=5/2.
Since f(5/2) = 13/6, the inflection point point is (5/2,13/6).
On the graph below, the two green points are the local maximum
and minimum, and the red point between is the inflection point.
Edwin
