Math / Calculus

QuestionLet $f$ be the function given by $f(x)=x+\tan \left(\frac{x}{5}\right)-10$ . The Intermediate Value Theorem applied to $f$ on the closed interval $[12,15]$ guarantees a solution in $[12,15]$ to which of the following equations? (A) $f(x)=-10$ (B) $f(x)=0$ (C) $f(x)=4$ D. $f(x)=14$

Studdy Solution

STEP 1

What is this asking? This problem is asking us to figure out which equation has a solution between $x = 12$ and $x = 15$ using the Intermediate Value Theorem applied to the function $f(x)$ . Watch out! The Intermediate Value Theorem doesn't tell us what the solution is, just that one exists!
Also, make sure you understand what the theorem actually says; it's not just about any value between $f(12)$ and $f(15)$ .

STEP 2

1. Evaluate f(12)
2. Evaluate f(15)
3. Apply the Intermediate Value Theorem

STEP 3

Let's plug in $x = 12$ into our function $f(x) = x + \tan\left(\frac{x}{5}\right) - 10$ .
So we have $f(12) = 12 + \tan\left(\frac{12}{5}\right) - 10$ .

STEP 4

This simplifies to $f(12) = 2 + \tan(2.4)$ .

STEP 5

Using a calculator, we find that $\tan(2.4) \approx -0.916$ .

STEP 6

Therefore, $f(12) \approx 2 + (-0.916) = 1.084$ .
This is our first key value!

STEP 7

Now let's plug in $x = 15$ into our function: $f(15) = 15 + \tan\left(\frac{15}{5}\right) - 10$ .

STEP 8

This simplifies to $f(15) = 5 + \tan(3)$ .

STEP 9

Using a calculator, we find that $\tan(3) \approx -0.1425$ .

STEP 10

Therefore, $f(15) \approx 5 + (-0.1425) = 4.8575$ .
This is our second key value!

STEP 11

The Intermediate Value Theorem says that if a function is continuous on a closed interval $[a, b]$ , then it takes on every value between $f(a)$ and $f(b)$ at some point within that interval.

STEP 12

The function $f(x)$ is continuous on the interval $[12, 15]$ since $x$ is a polynomial and $\tan\left(\frac{x}{5}\right)$ is continuous on this interval.

STEP 13

We found that $f(12) \approx 1.084$ and $f(15) \approx 4.8575$ .
So, $f(x)$ must take on every value between approximately $1.084$ and $4.8575$ on the interval $[12, 15]$ .

STEP 14

Since $0$ is not between $1.084$ and $4.8575$ , $f(x) = 0$ is not guaranteed a solution.
Similarly, $14$ is outside this range.
However, $4$ is between $1.084$ and $4.8575$ .
Therefore, the Intermediate Value Theorem guarantees a solution to $f(x) = 4$ in the interval $[12, 15]$ .

STEP 15

The Intermediate Value Theorem guarantees a solution to $f(x) = 4$ in the interval $[12, 15]$ , so the answer is (C).

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QuestionLet $f$ be the function given by $f(x)=x+\tan \left(\frac{x}{5}\right)-10$ . The Intermediate Value Theorem applied to $f$ on the closed interval $[12,15]$ guarantees a solution in $[12,15]$ to which of the following equations? (A) $f(x)=-10$ (B) $f(x)=0$ (C) $f(x)=4$ D. $f(x)=14$

Studdy Solution

STEP 1

STEP 2

1. Evaluate f(12)
2. Evaluate f(15)
3. Apply the Intermediate Value Theorem

STEP 3

Let's plug in $x = 12$ into our function $f(x) = x + \tan\left(\frac{x}{5}\right) - 10$ .
So we have $f(12) = 12 + \tan\left(\frac{12}{5}\right) - 10$ .

STEP 4

This simplifies to $f(12) = 2 + \tan(2.4)$ .

STEP 5

Using a calculator, we find that $\tan(2.4) \approx -0.916$ .

STEP 6

Therefore, $f(12) \approx 2 + (-0.916) = 1.084$ .
This is our first key value!

STEP 7

Now let's plug in $x = 15$ into our function: $f(15) = 15 + \tan\left(\frac{15}{5}\right) - 10$ .

STEP 8

This simplifies to $f(15) = 5 + \tan(3)$ .

STEP 9

Using a calculator, we find that $\tan(3) \approx -0.1425$ .

STEP 10

Therefore, $f(15) \approx 5 + (-0.1425) = 4.8575$ .
This is our second key value!

STEP 11

The Intermediate Value Theorem says that if a function is continuous on a closed interval $[a, b]$ , then it takes on every value between $f(a)$ and $f(b)$ at some point within that interval.

STEP 12

The function $f(x)$ is continuous on the interval $[12, 15]$ since $x$ is a polynomial and $\tan\left(\frac{x}{5}\right)$ is continuous on this interval.

STEP 13

We found that $f(12) \approx 1.084$ and $f(15) \approx 4.8575$ .
So, $f(x)$ must take on every value between approximately $1.084$ and $4.8575$ on the interval $[12, 15]$ .

STEP 14

Since $0$ is not between $1.084$ and $4.8575$ , $f(x) = 0$ is not guaranteed a solution.
Similarly, $14$ is outside this range.
However, $4$ is between $1.084$ and $4.8575$ .
Therefore, the Intermediate Value Theorem guarantees a solution to $f(x) = 4$ in the interval $[12, 15]$ .

STEP 15

The Intermediate Value Theorem guarantees a solution to $f(x) = 4$ in the interval $[12, 15]$ , so the answer is (C).

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Studdy Solution

STEP 1

STEP 2

1. Evaluate f(12)2. Evaluate f(15)3. Apply the Intermediate Value Theorem

STEP 3

Let's **plug in** x=12x = 12x=12 into our function f(x)=x+tan⁡(x5)−10f(x) = x + \tan\left(\frac{x}{5}\right) - 10f(x)=x+tan(5x​)−10.So we have f(12)=12+tan⁡(125)−10f(12) = 12 + \tan\left(\frac{12}{5}\right) - 10f(12)=12+tan(512​)−10.

STEP 4

This simplifies to f(12)=2+tan⁡(2.4)f(12) = 2 + \tan(2.4)f(12)=2+tan(2.4).

STEP 5

Using a calculator, we find that tan⁡(2.4)≈−0.916\tan(2.4) \approx -0.916tan(2.4)≈−0.916.

STEP 6

Therefore, f(12)≈2+(−0.916)=1.084f(12) \approx 2 + (-0.916) = 1.084f(12)≈2+(−0.916)=1.084.This is our **first key value**!

STEP 7

Now let's **plug in** x=15x = 15x=15 into our function: f(15)=15+tan⁡(155)−10f(15) = 15 + \tan\left(\frac{15}{5}\right) - 10f(15)=15+tan(515​)−10.

STEP 8

This simplifies to f(15)=5+tan⁡(3)f(15) = 5 + \tan(3)f(15)=5+tan(3).

STEP 9

Using a calculator, we find that tan⁡(3)≈−0.1425\tan(3) \approx -0.1425tan(3)≈−0.1425.

STEP 10

Therefore, f(15)≈5+(−0.1425)=4.8575f(15) \approx 5 + (-0.1425) = 4.8575f(15)≈5+(−0.1425)=4.8575.This is our **second key value**!

STEP 11

The Intermediate Value Theorem says that if a function is continuous on a closed interval [a,b][a, b][a,b], then it takes on every value between f(a)f(a)f(a) and f(b)f(b)f(b) at some point within that interval.

STEP 12

The function f(x)f(x)f(x) is continuous on the interval [12,15][12, 15][12,15] since xxx is a polynomial and tan⁡(x5)\tan\left(\frac{x}{5}\right)tan(5x​) is continuous on this interval.

STEP 13

We found that f(12)≈1.084f(12) \approx 1.084f(12)≈1.084 and f(15)≈4.8575f(15) \approx 4.8575f(15)≈4.8575.So, f(x)f(x)f(x) must take on every value between **approximately** 1.0841.0841.084 and 4.85754.85754.8575 on the interval [12,15][12, 15][12,15].

STEP 14

STEP 15

The Intermediate Value Theorem guarantees a solution to f(x)=4f(x) = 4f(x)=4 in the interval [12,15][12, 15][12,15], so the answer is (C).

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1. Evaluate f(12)
2. Evaluate f(15)
3. Apply the Intermediate Value Theorem

Let's plug in $x = 12$ into our function $f(x) = x + \tan\left(\frac{x}{5}\right) - 10$ .
So we have $f(12) = 12 + \tan\left(\frac{12}{5}\right) - 10$ .

This simplifies to $f(12) = 2 + \tan(2.4)$ .

Using a calculator, we find that $\tan(2.4) \approx -0.916$ .

Therefore, $f(12) \approx 2 + (-0.916) = 1.084$ .
This is our first key value!

Now let's plug in $x = 15$ into our function: $f(15) = 15 + \tan\left(\frac{15}{5}\right) - 10$ .

This simplifies to $f(15) = 5 + \tan(3)$ .

Using a calculator, we find that $\tan(3) \approx -0.1425$ .

Therefore, $f(15) \approx 5 + (-0.1425) = 4.8575$ .
This is our second key value!

The Intermediate Value Theorem says that if a function is continuous on a closed interval $[a, b]$ , then it takes on every value between $f(a)$ and $f(b)$ at some point within that interval.

The function $f(x)$ is continuous on the interval $[12, 15]$ since $x$ is a polynomial and $\tan\left(\frac{x}{5}\right)$ is continuous on this interval.

We found that $f(12) \approx 1.084$ and $f(15) \approx 4.8575$ .
So, $f(x)$ must take on every value between approximately $1.084$ and $4.8575$ on the interval $[12, 15]$ .

The Intermediate Value Theorem guarantees a solution to $f(x) = 4$ in the interval $[12, 15]$ , so the answer is (C).