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2. [-/3 Points] DETAILS MY NOTES Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval. $x^4 + x - 3 = 0$, $(1, 2)$ $f(x) = x^4 + x - 3$ is \textbf{Select} on the closed interval $[1, 2]$. $f(1) = \boxed{\text{ }}$ and $f(2) = \boxed{\text{}}$. Since $-1 < \boxed{\text{ }} < 15$, there is a number $c$ in $(1, 2)$ such that $f(c) = \boxed{\text{}}$ by the Intermediate Value Theorem. Thus, there is a \textbf{Select} of the equation $x^4 + x - 3 = 0$ in the interval $(1, 2)$. Submit Answer 3. [1/1 Points] DETAILS PREVIOUS ANSWERS MY NOTES

          2. [-/3 Points]
DETAILS
MY NOTES
Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.
$x^4 + x - 3 = 0$, $(1, 2)$
$f(x) = x^4 + x - 3$ is \textbf{Select} on the closed interval $[1, 2]$. $f(1) = \boxed{\text{ }}$ and $f(2) = \boxed{\text{}}$. Since $-1 < \boxed{\text{ }} < 15$, there is a number $c$ in $(1, 2)$ such that $f(c) = \boxed{\text{}}$ by the Intermediate Value Theorem. Thus, there is a \textbf{Select} of the equation $x^4 + x - 3 = 0$ in the interval $(1, 2)$.
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3. [1/1 Points] DETAILS PREVIOUS ANSWERS MY NOTES

2. [-/3 Points]
DETAILS
MY NOTES
Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.
x^4 + x - 3 = 0, (1, 2)
f(x) = x^4 + x - 3 is Select on the closed interval [1, 2]. f(1) = and f(2) =. Since -1 < < 15, there is a number c in (1, 2) such that f(c) = by the Intermediate Value Theorem. Thus, there is a Select of the equation x^4 + x - 3 = 0 in the interval (1, 2).
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3. [1/1 Points] DETAILS PREVIOUS ANSWERS MY NOTES

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