Determine the intervals on which the function is concave up or concave down and find the points of inflection.
f(x) = (x - 4)(5 - x3)
The second derivative Test for Concavity states that when f''(x) exists for all x in an open interval:
If f"(x) > 0, then f(x) is in that interval.
If f"(x) < 0, then f(x) is in that interval.
A point P = (c, f(c)) is a point of inflection of f(x) if the concavity changes from up to down or from down to up at x = c. Thus, we need to find where f''(c) is so that we can then determine any change in concavity at x = c.
Use c = 0 and c = 2 to divide the real line into three intervals where we will determine the sign of f"(x).
(-∞,0), (0,2), (2,∞)
Let's pick x1 = −1 in the first interval, x2 = 1 in the second interval, and x3 = 3 in the third interval. Fill in the table below with the appropriate sign.
Interval | x-value | Sign of f''(x) |
---|---|---|
(-∞,0) | -1 | |
(0,2) | 1 | |
(0,∞) | 3 |
Based on the sign of f''(x) in the table:
f(x) is on the interval (0,2).
f(x) is on the intervals (-∞,0) (2,∞).