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The second derivative of f 1x2
is
f –1x2 2 ˛x 5 3
9
2
9x 5 3
Since x 5 if and x 5 3
7 0 x 7 0,
3
6 0 if x 6 0, we obtain the following table:
Interval x 6 0 x 0 x 7 0
f (x)
f(x)
does not exist
The graph has a point of inflection when x 0, even though f ¿102 and f –102 do
not exist. Note that the curve crosses its tangent at x 0.
y
4
2
x
–6 –4 –2 0 2 4 6
–2
–4
EXAMPLE 4
Reasoning about points of inflection
Determine any points of inflection on the graph of f 1x2 1 .
x 2 3
Solution
The derivative of f 1x2 1 1x is f ¿1x2 2x 1x 2 32 2 .
x 2 3 2 32 1
The second derivative is
f –1x2 21x 2 32 2 4x1x 2 32 3 12x2
2
1x 2 32 2
21x2 32 8x 2
1x 2 32 3
6x2 6
1x 2 32 3
8x2
1x 2 32 3
NEL
CHAPTER 4 203