Question Find the interval where $x + \sqrt[3]{x} = 1$ has a solution, per the Intermediate Value Theorem.

Studdy Solution

STEP 1

Assumptions
1. We are working with the function $f(x) = x + \sqrt[3]{x}$ .
2. We want to find an interval where there is at least one solution to $f(x) = 1$ .
3. We will use the Intermediate Value Theorem, which states that if $f(x)$ is continuous on the interval $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$ , then there exists at least one number $c$ in the interval $(a, b)$ such that $f(c) = N$ .

STEP 2

First, we need to evaluate the function at the endpoints of each interval to find where the function value changes from being less than 1 to greater than 1 or vice versa.

STEP 3

Evaluate the function at the lower endpoint of the first interval $(0,1)$ , which is $f(0)$ .
$f(0) = 0 + \sqrt[3]{0} = 0$

STEP 4

Evaluate the function at the upper endpoint of the first interval $(0,1)$ , which is $f(1)$ .
$f(1) = 1 + \sqrt[3]{1} = 1 + 1 = 2$

STEP 5

Since $f(0) < 1 < f(1)$ , by the Intermediate Value Theorem, there is at least one solution to $f(x) = 1$ in the interval $(0,1)$ .

STEP 6

For completeness, let's check the other intervals to ensure that there are no other intervals where the solution could also exist.

STEP 7

Evaluate the function at the lower endpoint of the second interval $(1,2)$ , which is $f(1)$ .
$f(1) = 1 + \sqrt[3]{1} = 2$

STEP 8

Evaluate the function at the upper endpoint of the second interval $(1,2)$ , which is $f(2)$ .
$f(2) = 2 + \sqrt[3]{2} > 2$

STEP 9

Since $f(1) > 1$ and $f(2) > 1$ , there is no change in sign over the interval $(1,2)$ , so there is no solution to $f(x) = 1$ in this interval according to the Intermediate Value Theorem.

STEP 10

Evaluate the function at the lower endpoint of the third interval $(2,3)$ , which is $f(2)$ .
$f(2) = 2 + \sqrt[3]{2} > 2$

STEP 11

Evaluate the function at the upper endpoint of the third interval $(2,3)$ , which is $f(3)$ .
$f(3) = 3 + \sqrt[3]{3} > 3$

STEP 12

Since $f(2) > 1$ and $f(3) > 1$ , there is no change in sign over the interval $(2,3)$ , so there is no solution to $f(x) = 1$ in this interval according to the Intermediate Value Theorem.

STEP 13

Evaluate the function at the lower endpoint of the fourth interval $(3,4)$ , which is $f(3)$ .
$f(3) = 3 + \sqrt[3]{3} > 3$

STEP 14

Evaluate the function at the upper endpoint of the fourth interval $(3,4)$ , which is $f(4)$ .
$f(4) = 4 + \sqrt[3]{4} > 4$

STEP 15

Since $f(3) > 1$ and $f(4) > 1$ , there is no change in sign over the interval $(3,4)$ , so there is no solution to $f(x) = 1$ in this interval according to the Intermediate Value Theorem.

STEP 16

The only interval where there is a change in sign from below 1 to above 1 is the interval $(0,1)$ . Therefore, according to the Intermediate Value Theorem, there is at least one solution for $x + \sqrt[3]{x} = 1$ in the interval $(0,1)$ .
The solution is the interval $(0,1)$ .

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