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Problems
If
f
(
x
)
=
{
3
x
2
+
12
x
−
1
,
−
1
≤
x
≤
2
37
−
x
,
2
<
x
≤
3
f(x) = \begin{cases} 3x^2 + 12x - 1, & -1 \le x \le 2 \\ 37 - x, & 2 < x \le 3 \end{cases}
f
(
x
)
=
{
3
x
2
+
12
x
−
1
,
37
−
x
,
−
1
≤
x
≤
2
2
<
x
≤
3
…
Problem
If
f
(
x
)
=
{
3
x
2
+
12
x
−
1
,
−
1
≤
x
≤
2
37
−
x
,
2
<
x
≤
3
f(x) = \begin{cases} 3x^2 + 12x - 1, & -1 \le x \le 2 \\ 37 - x, & 2 < x \le 3 \end{cases}
f
(
x
)
=
{
3
x
2
+
12
x
−
1
,
37
−
x
,
−
1
≤
x
≤
2
2
<
x
≤
3
, then a.
f
(
x
)
f(x)
f
(
x
)
is increasing on
[
−
1
,
2
]
[-1, 2]
[
−
1
,
2
]
b.
f
(
x
)
f(x)
f
(
x
)
is continuous on
[
−
1
,
3
]
[-1, 3]
[
−
1
,
3
]
c.
f
′
(
x
)
f'(x)
f
′
(
x
)
does not exist at
x
=
2
x = 2
x
=
2
d.
f
(
x
)
f(x)
f
(
x
)
has the maximum value at
x
=
2
x = 2
x
=
2
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