
7-26 Calculus – Differentiation and Integration
Testing all the intervals to the left and right of (x) = 2 for f ′′(x) = 6x - 12, we see that f ′′(x) < 0 for
all x in (-¥, 2) and f ′′(x) > 0 for all x in (2, +¥).
Therefore, f is concave downwards on (-¥, 2) and concave upwards on (2, +¥). And f has a point
of inflection at (2, f(2)) = (2, -38).
Example 7.19 Investigate the curve (y - 2)
3
= x + 4 for the points of inflection.
Solution: We have y x= + +2 4
1 3
( ) .
/
Thus, and f ' x
dy
dx
x f '' x x( ) ( ) ( ) ( ) .
/ /
= = + = − +
− −
1
3
4
2
9
4
2 3 5 3
f '' x x
x
( ) ( )
.
/
= ⇒ − + =
⇒ = −
−
0
2
9
4 0
4
5 3
Testing all the intervals to the left and right of
x f '' ...