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Let f(x) = x⁵ + x³ + x + 1. Which statement correctly… | StemPy Lab
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Let f(x) = x⁵ + x³ + x + 1. Which statement correctly shows that f(x) = 0 has exactly one real root?
Mathematics
Hard
High School (11–12)
Functions and Calculus › Application of Derivatives
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Created by
StemPy Lab
Let
f
(
x
)
=
x
5
+
x
3
+
x
+
1
f(x) = x^5 + x^3 + x + 1
f
(
x
)
=
x
5
+
x
3
+
x
+
1
. Which statement correctly shows that
f
(
x
)
=
0
f(x) = 0
f
(
x
)
=
0
has exactly one real root?
f
f
f
has odd degree, so it must have exactly one real root.
f
(
0
)
=
1
>
0
f(0) = 1 > 0
f
(
0
)
=
1
>
0
and
f
f
f
is increasing, so the root is positive.
f
′
(
x
)
=
0
f'(x) = 0
f
′
(
x
)
=
0
has no real solution, so
f
(
x
)
=
0
f(x) = 0
f
(
x
)
=
0
has no real root.
f
′
(
x
)
=
5
x
4
+
3
x
2
+
1
>
0
f'(x) = 5x^4 + 3x^2 + 1 > 0
f
′
(
x
)
=
5
x
4
+
3
x
2
+
1
>
0
, so
f
f
f
is strictly increasing. Also
f
(
−
1
)
=
−
2
<
0
<
1
=
f
(
0
)
f(-1) = -2 < 0 < 1 = f(0)
f
(
−
1
)
=
−
2
<
0
<
1
=
f
(
0
)
.
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