a. Determine whether the Mean Value Theorem applies to the function on the interval .
b. If so, find the point(s) that are guaranteed to exist by the Mean Value Theorem.
a. Choose the correct answer below.
A. No, because the function is continuous on the interval , but is not differentiable on the interval .
B. No, because the function is differentiable on the interval , but is not continuous on the interval .
C. Yes, because the function is continuous on the interval and differentiable on the interval .
D. No, because the function is not continuous on the interval , and is not differentiable on the interval .

Ask by Garza Pena. in the United States

Mar 28,2025

Question

a. Determine whether the Mean Value Theorem applies to the function on the interval .
b. If so, find the point(s) that are guaranteed to exist by the Mean Value Theorem.
a. Choose the correct answer below.
A. No, because the function is continuous on the interval , but is not differentiable on the interval .
B. No, because the function is differentiable on the interval , but is not continuous on the interval .
C. Yes, because the function is continuous on the interval and differentiable on the interval .
D. No, because the function is not continuous on the interval , and is not differentiable on the interval .

Ask by Garza Pena. in the United States

Mar 28,2025

UpStudy AI · Accepted Answer

**Step 1. Check the conditions of the Mean Value Theorem (MVT):**

The MVT requires that:
- $$ f(x) $$ is continuous on the closed interval $$[a,b]$$,
- $$ f(x) $$ is differentiable on the open interval $$(a,b)$$.

Since $$ f(x)=4-x^{2} $$ is a polynomial, it is continuous and differentiable for all real $$ x $$. Therefore, it is continuous on $$[-1,2]$$ and differentiable on $$(-1,2)$$.

**Conclusion for part (a):**

The function meets the conditions for the Mean Value Theorem. The correct answer is:

**C. Yes, because the function is continuous on the interval $$[-1,2]$$ and differentiable on the interval $$(-1,2)$$.**

---

**Step 2. Apply the Mean Value Theorem to find the point $$ c $$ in $$(-1,2)$$:**

The MVT states there exists a $$ c $$ such that

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

where $$ a=-1 $$ and $$ b=2 $$.

**Step 2a. Calculate $$ f(a) $$ and $$ f(b) $$:**

$$
f(-1)=4-(-1)^2=4-1=3,
$$
$$
f(2)=4-2^2=4-4=0.
$$

**Step 2b. Compute the average rate of change:**

$$
\frac{f(2)-f(-1)}{2-(-1)}=\frac{0-3}{2+1}=\frac{-3}{3}=-1.
$$

So, the Mean Value Theorem guarantees there exists at least one point $$ c $$ in $$(-1,2)$$ such that

$$
f'(c)=-1.
$$

**Step 2c. Find $$ c $$ using the derivative of $$ f(x) $$:**

Differentiate $$ f(x) $$:

$$
f(x)=4-x^2 \quad \Longrightarrow \quad f'(x)=-2x.
$$

Set $$ f'(c)=-2c $$ equal to $$-1$$:

$$
-2c=-1.
$$

Solve for $$ c $$:

$$
c=\frac{1}{2}.
$$

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**Final Answers:**

- **(a)** The correct answer is **C**.
- **(b)** The point guaranteed by the MVT is $$ c=\frac{1}{2} $$.
**(a) C. Yes, because the function is continuous on the interval $$[-1,2]$$ and differentiable on the interval $$(-1,2)$$.** **(b) The point guaranteed by the Mean Value Theorem is $$ c=\frac{1}{2} $$.**

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