हनरजवरडव़ डरव़पनहपबडनजबुबलक़वक़वकड
Answers
Answer:
f
(
x
)
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
=
x
3
−
6
x
2
+
11
x
−
6
is a cubic with positive leading coefficient and zeros of multiplicity
1
at
x
=
1
,
x
=
2
and
x
=
3
.
As a result:
f
(
x
)
is continuous.
f
(
x
)
is positive for large positive values of
x
.
f
(
x
)
is negative for large negative values of
x
.
f
(
x
)
changes sign at each of its zeros.
Hence
f
(
x
)
is non-negative in the intervals
[
1
,
2
]
and
[
3
,
∞
)
graph{x^3-6x^2+11x-6 [-3.397, 6.603, -2.24, 2.76]}
Answer:
f
(
x
)
=
(
x
−
1
)
(
x
−
2
)
(
x
−
3
)
=
x
3
−
6
x
2
+
11
x
−
6
is a cubic with positive leading coefficient and zeros of multiplicity
1
at
x
=
1
,
x
=
2
and
x
=
3
.
As a result:
f
(
x
)
is continuous.
f
(
x
)
is positive for large positive values of
x
.
f
(
x
)
is negative for large negative values of
x
.
f
(
x
)
changes sign at each of its zeros.
Hence
f
(
x
)
is non-negative in the intervals
[
1
,
2
]
and
[
3
,
∞
)
graph{x^3-6x^2+11x-6 [-3.397, 6.603, -2.24, 2.76]}