Answer:
1. f is increasing over the interval 0 < x < 2.
2. f(2) = 3
3. f is decreasing over the interval 2 < x < 5.
|DF| = |DE| + |EF| |DF| = 9x -36 |DE| = 47 |EF| = 3x+10 Substitute: 9x - 39 = 47 + 3x + 10 9x - 39 = 3x + 57 |+39 9x = 3x + 96 |-3x 6x = 96 |:6 x = 16 Put the value of x to the equation |EF| = 3x + 10 |EF| = (3)(16) + 10 = 48 + 10 = 58 Answer: |EF| = 58
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B i hope it helps if not sorry
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