(a) On what intervals is f increasing? (Select all that apply.) (0, 2) (0, 3) (0
ID: 2858831 • Letter: #
Question
(a) On what intervals is f increasing? (Select all that apply.)
(0, 2)
(0, 3)
(0, 6)
(2, 4)
(3, 6)
(4, 6)
(6, 8)
(6, 9)
(8, 9)
none of these
On what interval(s) is f decreasing? (Select all that apply.)
(0, 2)
(0, 3)
(0, 6)
(2, 4)
(3, 6)
(4, 6)
(6, 8)
(6, 9)
(8, 9)
none of these
(b) At what value(s) of x does f have a local maximum? (Select all that apply.)
1
2
3
4
5
6
7
8
none of these
At what value(s) of x does f have a local minimum? (Select all that apply.)
1
2
3
4
5
6
7
8
none of these
(c) On what interval(s) is f concave upward? (Select all that apply.)
(0, 2)
(0, 3)
(0, 6)
(2, 4)
(3, 6)
(2, 6)
(6, 9)
(5, 9)
(0, 9)
none of these
On what interval(s) is f concave downward? (Select all that apply.)
(0, 4)
(0, 3)
(0, 6)
(2, 4)
(3, 6)
(4, 6)
(6, 9)
(8, 9)
none of these
(d) What are the x-coordinate(s) of the inflection point(s) of f ? (Select all that apply.)
1
2
3
4
5
6
7
8
none of these
y-f'(x) 0Explanation / Answer
f'(x)>0for (0,2) U ( 4,6) U (8,9 ) , hence f will be increasing on these intervals.
f'(x) < 0 for ( 2, 4) U ( 6, 8) Hence f will be decreasing over this interval.
Since derivative is changing sign from +ve to -ve at x=2 , Hence x=2 is point of maxima.
Since derivative is changing sign from -ve to +ve at x=4, x=8, Hence f will be minimum at these values.
Slope of f'(x) is positive for ( 3 , 6 ) U (6 , 9 ) , Hence concave up
Slope of f'(x) is negative for ( 0 , 3) , Hence concave down.
Derivative of f'(x) is 0 at x = 3 and undefined at x= 6 , Hence 3, 6 are points of infelction.
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