Question Taylor uses Intermediate Value Theorem to solve $x^3 - 3x^2 - 9 = K$ over $[3, 9]$ . What is the possible value of $K$ ?

Studdy Solution

STEP 1

Assumptions
1. We are using the Intermediate Value Theorem (IVT), which states that if a function $f(x)$ is continuous on a closed 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 $f(c) = N$ .
2. The function given is $f(x) = x^3 - 3x^2 - 9$ .
3. The interval over which we are applying the IVT is $[3, 9]$ .
4. We are looking for the value of $K$ such that $f(x) = K$ has a solution in the interval $[3, 9]$ .

STEP 2

First, we need to evaluate the function at the endpoints of the interval to determine the range of values it takes on the interval. We will start by evaluating $f(3)$ .
$f(3) = 3^3 - 3(3)^2 - 9$

STEP 3

Calculate the value of $f(3)$ .
$f(3) = 27 - 3(9) - 9$ $f(3) = 27 - 27 - 9$ $f(3) = -9$

STEP 4

Next, we evaluate the function at the other endpoint of the interval, $f(9)$ .
$f(9) = 9^3 - 3(9)^2 - 9$

STEP 5

Calculate the value of $f(9)$ .
$f(9) = 729 - 3(81) - 9$ $f(9) = 729 - 243 - 9$ $f(9) = 486 - 9$ $f(9) = 477$

STEP 6

Now that we have the values of the function at the endpoints of the interval, we can apply the Intermediate Value Theorem. The IVT tells us that if $K$ is between $f(3)$ and $f(9)$ , then there is a solution to the equation $f(x) = K$ in the interval $[3, 9]$ .

STEP 7

Determine the range of $K$ values for which there is a guaranteed solution based on the IVT.
Since $f(3) = -9$ and $f(9) = 477$ , the range of $K$ values for which there is a guaranteed solution is $-9 \leq K \leq 477$ .

STEP 8

Now we will compare the given values of $K$ to the range determined in STEP_7 to see which, if any, could be the value of $K$ for which there is a solution to $f(x) = K$ in the interval $[3, 9]$ .

STEP 9

Check if $K = 702$ is in the range $-9 \leq K \leq 477$ .
$702$ is not between $-9$ and $477$ , so $K = 702$ cannot be the value of $K$ .

STEP 10

Check if $K = -10$ is in the range $-9 \leq K \leq 477$ .
$-10$ is not between $-9$ and $477$ , so $K = -10$ cannot be the value of $K$ .

STEP 11

Check if $K = 486$ is in the range $-9 \leq K \leq 477$ .
$486$ is not between $-9$ and $477$ , so $K = 486$ cannot be the value of $K$ .

STEP 12

Check if $K = -8$ is in the range $-9 \leq K \leq 477$ .
$-8$ is between $-9$ and $477$ , so $K = -8$ could be the value of $K$ .

STEP 13

Check if $K = -11$ is in the range $-9 \leq K \leq 477$ .
$-11$ is not between $-9$ and $477$ , so $K = -11$ cannot be the value of $K$ .

STEP 14

Based on the checks in the previous steps, the only value of $K$ that could be the solution to the equation $f(x) = K$ in the interval $[3, 9]$ according to the Intermediate Value Theorem is $K = -8$ .
The value of $K$ is $-8$ .

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