QuestionFind $x$ such that $f(x)=0$ for $f(x)=x^{3}-3x+1$ using the Intermediate Value Theorem.

Studdy Solution

STEP 1

Assumptions1. The function is $f(x)=x^{3}-3 x+1$ . The Intermediate Value Theorem (IV) states that if a continuous function, f, with an interval, [a, b], as its domain, takes values f(a) and f(b) at each end of the interval, then it also takes any value between f(a) and f(b) at some point within the interval.

STEP 2

First, we need to find the values of the function at some points to determine an interval where the function crosses the x-axis (i.e., where $f(x)=0$ ). Let's start by calculating $f(-2)$ and $f(-1)$ .
$f(-2) = (-2)^{}-(-2)+1$ $f(-1) = (-1)^{}-(-1)+1$

STEP 3

Calculate the values of $f(-2)$ and $f(-1)$ .
$f(-2) = -8+6+1 = -1$ $f(-1) = -1+3+1 =3$

STEP 4

From the calculations in3, we see that $f(-2) <0$ and $f(-1) >0$ . According to the Intermediate Value Theorem, since the function is continuous and changes sign between $x=-2$ and $x=-1$ , there must be at least one root in the interval $[-2,-1]$ .
Therefore, the Intermediate Value Theorem implies that $f(x)=0$ for some $x$ satisfying $-2 \leq x \leq -1$ .

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