Question

1. -/6 points use the intermediate value theorem to show that there is a root of the given equation in the specified interval. (x^{4}+x - 6=0), ((1,2)) (f(x)=x^{4}+x - 6) is select on the closed interval (1,2), (f(1)=) , and (f(2)=) . since (-4<?<12), 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 - 6 = 0) in the interval ((1,2)).

Explanation:

Step1: Recall continuity of polynomial

The function $f(x)=x^{4}+x - 6$ is a polynomial, and polynomials are continuous everywhere, so it is continuous on the closed - interval $[1,2]$.

Step2: Evaluate $f(1)$

Substitute $x = 1$ into $f(x)$:
$f(1)=1^{4}+1 - 6=1 + 1-6=-4$.

Step3: Evaluate $f(2)$

Substitute $x = 2$ into $f(x)$:
$f(2)=2^{4}+2 - 6=16 + 2-6=12$.

Step4: Apply Intermediate - Value Theorem

Since $f(x)$ is continuous on $[1,2]$, and $f(1)=-4$ and $f(2)=12$, and $0$ is between $-4$ and $12$ (i.e., $-4<0<12$), by the Intermediate - Value Theorem, there exists a number $c\in(1,2)$ such that $f(c) = 0$. So there is a root of the equation $x^{4}+x - 6=0$ in the interval $(1,2)$.

Answer:

$f(x)=x^{4}+x - 6$ is continuous on the closed interval $[1,2]$, $f(1)=-4$, and $f(2)=12$. Since $-4 < 0<12$, there is a number $c\in(1,2)$ such that $f(c)=0$ by the Intermediate - Value Theorem. Thus, there is a root of the equation $x^{4}+x - 6 = 0$ in the interval $(1,2)$.