Math / Calculus

QuestionDetermine over what interval(s) (if any) the Mean Value Theorem applies. (Enter your answer using interval notation. If an answer does not exist, enter DNE.) $y=\frac{1}{x^{3}}$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the intervals where the Mean Value Theorem (MVT) can be used with the function $y = \frac{1}{x^3}$ . Watch out! The MVT has two key conditions!
Don't forget to check both!

STEP 2

1. Check for continuity
2. Check for differentiability
3. Find the intervals

STEP 3

Let's investigate the function $y = \frac{1}{x^3}$ .
Is it continuous everywhere?
Nope! There's a big, giant discontinuity at $x = 0$ .
Dividing by zero is a no-no!

STEP 4

Now, let's think about where our function is differentiable.
Well, since $y = \frac{1}{x^3}$ can be rewritten as $y = x^{-3}$ , we can easily find its derivative: $\frac{dy}{dx} = -3x^{-4}$ , or $\frac{dy}{dx} = \frac{-3}{x^4}$ .

STEP 5

Just like with continuity, this derivative has a problem at $x = 0$ .
So, our function is not differentiable at $x = 0$ .

STEP 6

The Mean Value Theorem applies when the function is continuous on a closed interval, $[a, b]$ , and differentiable on the open interval, $(a, b)$ .

STEP 7

We know our function is smooth and happy everywhere except at $x = 0$ .
So, the MVT applies on any interval that doesn't include zero!

STEP 8

That means the MVT applies on $(-\infty, 0)$ and $(0, \infty)$ .

STEP 9

The Mean Value Theorem applies on the intervals $(-\infty, 0)$ and $(0, \infty)$ .

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QuestionDetermine over what interval(s) (if any) the Mean Value Theorem applies. (Enter your answer using interval notation. If an answer does not exist, enter DNE.) $y=\frac{1}{x^{3}}$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the intervals where the Mean Value Theorem (MVT) can be used with the function $y = \frac{1}{x^3}$ . Watch out! The MVT has two key conditions!
Don't forget to check both!

STEP 2

1. Check for continuity
2. Check for differentiability
3. Find the intervals

STEP 3

Let's investigate the function $y = \frac{1}{x^3}$ .
Is it continuous everywhere?
Nope! There's a big, giant discontinuity at $x = 0$ .
Dividing by zero is a no-no!

STEP 4

Now, let's think about where our function is differentiable.
Well, since $y = \frac{1}{x^3}$ can be rewritten as $y = x^{-3}$ , we can easily find its derivative: $\frac{dy}{dx} = -3x^{-4}$ , or $\frac{dy}{dx} = \frac{-3}{x^4}$ .

STEP 5

Just like with continuity, this derivative has a problem at $x = 0$ .
So, our function is not differentiable at $x = 0$ .

STEP 6

The Mean Value Theorem applies when the function is continuous on a closed interval, $[a, b]$ , and differentiable on the open interval, $(a, b)$ .

STEP 7

We know our function is smooth and happy everywhere except at $x = 0$ .
So, the MVT applies on any interval that doesn't include zero!

STEP 8

That means the MVT applies on $(-\infty, 0)$ and $(0, \infty)$ .

STEP 9

The Mean Value Theorem applies on the intervals $(-\infty, 0)$ and $(0, \infty)$ .

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QuestionDetermine over what interval(s) (if any) the Mean Value Theorem applies. (Enter your answer using interval notation. If an answer does not exist, enter DNE.) y=1x3y=\frac{1}{x^{3}}y=x31​ □\square□

Studdy Solution

STEP 1

What is this asking? We need to find the intervals where the Mean Value Theorem (MVT) can be used with the function y=1x3y = \frac{1}{x^3}y=x31​. Watch out! The MVT has *two* key conditions!Don't forget to check both!

STEP 2

1. Check for continuity2. Check for differentiability3. Find the intervals

STEP 3

Let's **investigate** the function y=1x3y = \frac{1}{x^3}y=x31​.Is it continuous everywhere?Nope! There's a big, giant **discontinuity** at x=0x = 0x=0.Dividing by zero is a no-no!

STEP 4

STEP 5

Just like with continuity, this derivative has a **problem** at x=0x = 0x=0.So, our function is **not differentiable** at x=0x = 0x=0.

STEP 6

The Mean Value Theorem applies when the function is **continuous** on a **closed interval**, [a,b][a, b][a,b], and **differentiable** on the **open interval**, (a,b)(a, b)(a,b).

STEP 7

We know our function is **smooth and happy** everywhere *except* at x=0x = 0x=0.So, the MVT applies on any interval that *doesn't include* zero!

STEP 8

That means the MVT applies on (−∞,0)(-\infty, 0)(−∞,0) and (0,∞)(0, \infty)(0,∞).

STEP 9

The Mean Value Theorem applies on the intervals (−∞,0)(-\infty, 0)(−∞,0) and (0,∞)(0, \infty)(0,∞).

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionDetermine over what interval(s) (if any) the Mean Value Theorem applies. (Enter your answer using interval notation. If an answer does not exist, enter DNE.) $y=\frac{1}{x^{3}}$ $\square$

What is this asking? We need to find the intervals where the Mean Value Theorem (MVT) can be used with the function $y = \frac{1}{x^3}$ . Watch out! The MVT has two key conditions!
Don't forget to check both!

1. Check for continuity
2. Check for differentiability
3. Find the intervals

Let's investigate the function $y = \frac{1}{x^3}$ .
Is it continuous everywhere?
Nope! There's a big, giant discontinuity at $x = 0$ .
Dividing by zero is a no-no!

Just like with continuity, this derivative has a problem at $x = 0$ .
So, our function is not differentiable at $x = 0$ .

The Mean Value Theorem applies when the function is continuous on a closed interval, $[a, b]$ , and differentiable on the open interval, $(a, b)$ .

We know our function is smooth and happy everywhere except at $x = 0$ .
So, the MVT applies on any interval that doesn't include zero!

That means the MVT applies on $(-\infty, 0)$ and $(0, \infty)$ .

The Mean Value Theorem applies on the intervals $(-\infty, 0)$ and $(0, \infty)$ .