Question

1. (4 points) use the intermediate value theorem to show that the equation $x^{3}-x + 2=0$ has a solution between $x=-2$ and $x = 0$. to do this problem you must come to nac 1/511 and talk to david john.

Explanation:

Step1: Define the function

Let $f(x)=x^{3}-x + 2$.

Step2: Evaluate the function at the endpoints

When $x=-2$, $f(-2)=(-2)^{3}-(-2)+2=-8 + 2+2=-4$. When $x = 0$, $f(0)=0^{3}-0 + 2=2$.

Step3: Apply the Intermediate - Value Theorem

The Intermediate - Value Theorem states that if $y = f(x)$ is continuous on a closed interval $[a,b]$, and $k$ is a number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in the interval $(a,b)$ such that $f(c)=k$. Here, the function $y = f(x)=x^{3}-x + 2$ is a polynomial function, so it is continuous on the interval $[-2,0]$. The number $0$ is between $f(-2)=-4$ and $f(0)=2$. So there exists a number $c\in(-2,0)$ such that $f(c)=0$, which means the equation $x^{3}-x + 2=0$ has a solution in the interval $(-2,0)$.

Answer:

The equation $x^{3}-x + 2=0$ has a solution between $x=-2$ and $x = 0$ as shown by the Intermediate - Value Theorem.