Suppose
f
′
(
x
)
=
0
\displaystyle {f}'{\left({x}\right)}={0}
f
′
(
x
)
=
0
at
x
=
x
0
\displaystyle {x}={x}_{{{0}}}
x
=
x
0
but nowhere else. Also assume
f
′
(
x
)
\displaystyle {f}'{\left({x}\right)}
f
′
(
x
)
is continuous and has the following values, where
a
≤
x
0
≤
b
\displaystyle {a}\le{x}_{{{0}}}\le{b}
a
≤
x
0
≤
b
:
x
\displaystyle {x}
x
a
\displaystyle {a}
a
x
0
\displaystyle {x}_{{{0}}}
x
0
b
\displaystyle {b}
b
f
′
(
x
)
\displaystyle {f}'{\left({x}\right)}
f
′
(
x
)
−
4
\displaystyle -{4}
−
4
0
\displaystyle {0}
0
−
5
\displaystyle -{5}
−
5
Check ALL that apply:
f
\displaystyle {f}
f
increases after
x
=
x
0
\displaystyle {x}={x}_{{{0}}}
x
=
x
0
f
\displaystyle {f}
f
has a local minimum at
x
=
x
0
\displaystyle {x}={x}_{{{0}}}
x
=
x
0
f
\displaystyle {f}
f
decreases until
x
=
x
0
\displaystyle {x}={x}_{{{0}}}
x
=
x
0
f
\displaystyle {f}
f
decreases after
x
=
x
0
\displaystyle {x}={x}_{{{0}}}
x
=
x
0
f
\displaystyle {f}
f
has a local maximum at
x
=
x
0
\displaystyle {x}={x}_{{{0}}}
x
=
x
0
This means that:
f
′
(
x
0
)
=
0
\displaystyle {f}'{\left({x}_{{{0}}}\right)}={0}
f
′
(
x
0
)
=
0
guarantees neither a local maximum nor a local minimum
x
=
x
0
\displaystyle {x}={x}_{{{0}}}
x
=
x
0
If
f
′
(
x
0
)
=
0
\displaystyle {f}'{\left({x}_{{{0}}}\right)}={0}
f
′
(
x
0
)
=
0
, there is a local maximum at
x
=
x
0
\displaystyle {x}={x}_{{{0}}}
x
=
x
0
The given information is incorrect.
If
f
′
(
x
0
)
=
0
\displaystyle {f}'{\left({x}_{{{0}}}\right)}={0}
f
′
(
x
0
)
=
0
, there is a local minimum at
x
=
x
0
\displaystyle {x}={x}_{{{0}}}
x
=
x
0
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