Name: 14. For this problem, $$ f(x)=x^{2} $$. Because $$ f $$ is differentiable everywhere, it satisfies all hypotheses of the MVT on any interval $$ [a, b] $$. a) What does the MVT say about $$ f $$ on $$ [\mathrm{a}, \mathrm{b}]=[-1,1] $$ ? Find all suitable values of $$ c $$. How many are there? b) What does the MVT say about $$ f $$ on $$ [\mathrm{a}, \mathrm{b}]=[1,2] $$ ? Find all suitable values of $$ c $$. How many are there? c) Show that for $$ f(x)=x^{2} $$ and any interval $$ [a, b] $$, the MVT's number $$ c $$ is the midpoint of $$ [\mathrm{a}, \mathrm{b}] $$.

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Let's analyze each part of the problem step by step.

## Given:
- Function: $$ f(x) = x^2 $$
- The function $$ f $$ is differentiable everywhere, so it satisfies all hypotheses of the Mean Value Theorem (MVT) on any interval $$[a, b]$$.

---

## a) Applying the MVT on $$[-1, 1]$$

### **Mean Value Theorem Statement:**
There exists at least one $$ c $$ in $$(a, b)$$ such that:
$$
f'(c) = \frac{f(b) - f(a)}{b - a}
$$

### **Applying to $$[-1, 1]$$:**
1. **Compute $$ f(a) $$ and $$ f(b) $$:**
   $$
   f(-1) = (-1)^2 = 1 \
   f(1) = 1^2 = 1
   $$
   
2. **Compute the average rate of change:**
   $$
   \frac{f(1) - f(-1)}{1 - (-1)} = \frac{1 - 1}{2} = 0
   $$
   
3. **Set $$ f'(c) $$ equal to the average rate of change:**
   $$
   f'(c) = 2c = 0 \
   \Rightarrow c = 0
   $$
   
4. **Verify $$ c $$ is within $$(-1, 1)$$:**
   $$
   0 \in (-1, 1)
   $$
   
### **Conclusion:**
- **Suitable Value of $$ c $$:** $$ c = 0 $$
- **Number of Suitable Values:** **1**

---

## b) Applying the MVT on $$[1, 2]$$

### **Applying to $$[1, 2]$$:**
1. **Compute $$ f(a) $$ and $$ f(b) $$:**
   $$
   f(1) = 1^2 = 1 \
   f(2) = 2^2 = 4
   $$
   
2. **Compute the average rate of change:**
   $$
   \frac{f(2) - f(1)}{2 - 1} = \frac{4 - 1}{1} = 3
   $$
   
3. **Set $$ f'(c) $$ equal to the average rate of change:**
   $$
   f'(c) = 2c = 3 \
   \Rightarrow c = \frac{3}{2} = 1.5
   $$
   
4. **Verify $$ c $$ is within $$(1, 2)$$:**
   $$
   1.5 \in (1, 2)
   $$
   
### **Conclusion:**
- **Suitable Value of $$ c $$:** $$ c = 1.5 $$
- **Number of Suitable Values:** **1**

---

## c) Proving $$ c $$ is the Midpoint for Any Interval $$[a, b]$$

### **Objective:**
Show that for $$ f(x) = x^2 $$ and any interval $$[a, b]$$, the value of $$ c $$ guaranteed by the MVT is the midpoint of $$[a, b]$$, i.e., $$ c = \frac{a + b}{2} $$.

### **Proof:**
1. **Compute the average rate of change:**
   $$
   \frac{f(b) - f(a)}{b - a} = \frac{b^2 - a^2}{b - a} = \frac{(b - a)(b + a)}{b - a} = b + a
   $$
   
2. **Set $$ f'(c) $$ equal to the average rate of change:**
   $$
   f'(c) = 2c = a + b \
   \Rightarrow c = \frac{a + b}{2}
   $$
   
3. **Interpretation:**
   $$
   c = \frac{a + b}{2}
   $$
   This is exactly the formula for the midpoint of the interval $$[a, b]$$.

### **Conclusion:**
For any interval $$[a, b]$$, the value $$ c $$ guaranteed by the MVT for the function $$ f(x) = x^2 $$ is the **midpoint** of the interval, $$ c = \frac{a + b}{2} $$.

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For the function $$ f(x) = x^2 $$: a) On the interval $$[-1, 1]$$, the MVT says there's one value $$ c = 0 $$ where the slope equals the average rate of change. b) On the interval $$[1, 2]$$, the MVT says there's one value $$ c = 1.5 $$ where the slope equals the average rate of change. c) For any interval $$[a, b]$$, the MVT's value $$ c $$ is the midpoint $$ \frac{a + b}{2} $$.

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