Name: 10. Explain why it is not valid to use the Mean Value Theorem for these functions on the indicated interval. On which of these 4 problems is the conclusion of the Mean Value Theorem true even thought the hypotheses is false? $$ \begin{array}{ll} ext { a) } f(x)=\frac{1}{x},[-1,1] & ext { b) } f(x)=\frac{1}{x^{2}},[-1,2]\end{array} $$ $$ \begin{array}{ll} ext { c) } f(x)= an x,[0, \pi] & ext { d) } f(x)=x^{1 / 3},[-1,1]\end{array} $$

Question

Name: 10. Explain why it is not valid to use the Mean Value Theorem for these functions on the indicated interval. On which of these 4 problems is the conclusion of the Mean Value Theorem true even thought the hypotheses is false? $$ \begin{array}{ll}	ext { a) } f(x)=\frac{1}{x},[-1,1] & 	ext { b) } f(x)=\frac{1}{x^{2}},[-1,2]\end{array} $$ $$ \begin{array}{ll}	ext { c) } f(x)=	an x,[0, \pi] & 	ext { d) } f(x)=x^{1 / 3},[-1,1]\end{array} $$

UpStudy AI · Accepted Answer

To determine the validity of using the Mean Value Theorem (MVT) for the given functions on the specified intervals, we need to recall the conditions of the MVT. The MVT states that if a function $$ f $$ is continuous on the closed interval $$[a, b]$$ and differentiable on the open interval $$(a, b)$$, then there exists at least one point $$ c $$ in $$(a, b)$$ such that:

$$
f'(c) = \frac{f(b) - f(a)}{b - a}
$$

Now, let's analyze each function and interval:

### a) $$ f(x) = \frac{1}{x}, [-1, 1] $$

- **Continuity**: The function $$ f(x) = \frac{1}{x} $$ is not continuous on the interval $$[-1, 1]$$ because it is undefined at $$ x = 0 $$.
- **Differentiability**: Since the function is not continuous, it cannot be differentiable.
- **Conclusion**: The MVT cannot be applied because the function is not continuous on the interval.

### b) $$ f(x) = \frac{1}{x^2}, [-1, 2] $$

- **Continuity**: The function $$ f(x) = \frac{1}{x^2} $$ is not continuous on the interval $$[-1, 2]$$ because it is undefined at $$ x = 0 $$.
- **Differentiability**: Since the function is not continuous, it cannot be differentiable.
- **Conclusion**: The MVT cannot be applied because the function is not continuous on the interval.

### c) $$ f(x) = \tan x, [0, \pi] $$

- **Continuity**: The function $$ f(x) = \tan x $$ is continuous on the interval $$[0, \pi]$$ except at $$ x = \frac{\pi}{2} $$, where it is undefined.
- **Differentiability**: The function is not differentiable at $$ x = \frac{\pi}{2} $$.
- **Conclusion**: The MVT cannot be applied because the function is not continuous on the entire interval.

### d) $$ f(x) = x^{1/3}, [-1, 1] $$

- **Continuity**: The function $$ f(x) = x^{1/3} $$ is continuous on the interval $$[-1, 1]$$.
- **Differentiability**: The function is differentiable on the interval $$(-1, 1)$$.
- **Conclusion**: The MVT can be applied here, and it is valid.

### Summary of Validity

- **a)** Not valid (discontinuous at $$ x = 0 $$)
- **b)** Not valid (discontinuous at $$ x = 0 $$)
- **c)** Not valid (discontinuous at $$ x = \frac{\pi}{2} $$)
- **d)** Valid (continuous and differentiable)

### Conclusion of MVT Being True Despite Hypotheses Being False

The conclusion of the MVT can still be true even if the hypotheses are false. In this case, the function $$ f(x) = \tan x $$ on the interval $$[0, \pi]$$ does not satisfy the hypotheses of the MVT, but we can still find a point $$ c $$ where the conclusion holds true, as the function has a vertical asymptote at $$ x = \frac{\pi}{2} $$.

Thus, the answer is:
- **Functions where MVT cannot be applied**: a, b, c
- **Function where MVT can be applied**: d
- **Conclusion true despite hypotheses false**: c (for $$ f(x) = \tan x $$)
The Mean Value Theorem cannot be applied to functions a, b, and c because they are not continuous on their respective intervals. However, for function d ($$ f(x) = x^{1/3} $$ on $$[-1, 1]$$), the theorem can be applied even though the function is not differentiable at $$ x = 0 $$.

Name:
10. Explain why it is not valid to use the Mean Value Theorem for these functions on the indicated interval. On
which of these 4 problems is the conclusion of the Mean Value Theorem true even thought the hypotheses is
false?

Solución de inteligencia artificial de Upstudy

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Calculus Questions