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f
(
x
)
=
{
{
2
−
∣
x
2
+
5
x
+
6
∣
,
x
≠
−
2
a
2
+
1
,
x
=
−
2
f(x) = \left\{\begin{cases} 2 - |x^2 + 5x + 6|, & x \ne -2 \\ a^2 + 1, & x = -2 \end{cases}\right.
f
(
x
)
=
{
{
2
−
∣
x
2
+
5
x
+
6∣
,
a
2
+
1
,
x
=
−
2
x
=
−
2
…
Problem
f
(
x
)
=
{
{
2
−
∣
x
2
+
5
x
+
6
∣
,
x
≠
−
2
a
2
+
1
,
x
=
−
2
f(x) = \left\{\begin{cases} 2 - |x^2 + 5x + 6|, & x \ne -2 \\ a^2 + 1, & x = -2 \end{cases}\right.
f
(
x
)
=
{
{
2
−
∣
x
2
+
5
x
+
6∣
,
a
2
+
1
,
x
=
−
2
x
=
−
2
, then the range of
a
a
a
, so that
f
(
x
)
f(x)
f
(
x
)
has maxima at
x
=
−
2
x = -2
x
=
−
2
is a.
∣
a
∣
≥
1
|a| \ge 1
∣
a
∣
≥
1
b.
∣
a
∣
<
1
|a| < 1
∣
a
∣
<
1
c.
a
>
1
a > 1
a
>
1
d.
a
<
1
a < 1
a
<
1
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