Problem 8: Let $$ f(x)=3 x^{3}-2 x^{2}+x $$ on the interval $$ [-1,2] $$. . Apply the MVT to find the value(s) of $$ c $$ in $$ (-1,2) $$ where $$ f^{\prime}(c)=\frac{f(2)-f(-1)}{2-(-1)} $$.

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To apply the Mean Value Theorem (MVT) to the function $$ f(x) = 3x^3 - 2x^2 + x $$ on the interval $$[-1, 2]$$, follow these steps:

### **1. Verify the Conditions for MVT**

- **Continuity:** $$ f(x) $$ is a polynomial, so it is continuous on $$[-1, 2]$$.
- **Differentiability:** Since $$ f(x) $$ is a polynomial, it is differentiable on $$(-1, 2)$$.

### **2. Apply the Mean Value Theorem**

According to the MVT, there exists at least one $$ c $$ in $$(-1, 2)$$ such that:
$$
f'(c) = \frac{f(2) - f(-1)}{2 - (-1)}
$$

### **3. Compute $$ f(2) $$ and $$ f(-1) $$**

$$
f(2) = 3(2)^3 - 2(2)^2 + 2 = 24 - 8 + 2 = 18
$$
$$
f(-1) = 3(-1)^3 - 2(-1)^2 + (-1) = -3 - 2 - 1 = -6
$$

### **4. Calculate the Average Rate of Change**

$$
\frac{f(2) - f(-1)}{2 - (-1)} = \frac{18 - (-6)}{3} = \frac{24}{3} = 8
$$

### **5. Find $$ f'(x) $$ and Solve for $$ c $$**

First, find the derivative $$ f'(x) $$:
$$
f'(x) = 9x^2 - 4x + 1
$$
Set $$ f'(c) = 8 $$:
$$
9c^2 - 4c + 1 = 8 \
9c^2 - 4c - 7 = 0
$$

### **6. Solve the Quadratic Equation**

$$
c = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 9 \cdot (-7)}}{2 \cdot 9} = \frac{4 \pm \sqrt{16 + 252}}{18} = \frac{4 \pm \sqrt{268}}{18}
$$
Simplify the square root:
$$
\sqrt{268} = 2\sqrt{67}
$$
Thus,
$$
c = \frac{4 \pm 2\sqrt{67}}{18} = \frac{2 \pm \sqrt{67}}{9}
$$

### **7. Verify $$ c $$ is in $$(-1, 2)$$**

Calculate approximate values:
$$
\sqrt{67} \approx 8.185
$$
$$
c_1 = \frac{2 + 8.185}{9} \approx 1.132 \quad (\text{within } (-1, 2))
$$
$$
c_2 = \frac{2 - 8.185}{9} \approx -0.687 \quad (\text{within } (-1, 2))
$$

Both values of $$ c $$ lie within the interval $$(-1, 2)$$.

### **Final Answer**

The values of $$ c $$ in $$(-1, 2)$$ satisfying the Mean Value Theorem are:
$$
c = \frac{2 + \sqrt{67}}{9} \quad \text{and} \quad c = \frac{2 - \sqrt{67}}{9}
$$

**Answer:** After simplifying, the values c are (2 ± √67) divided by 9. Thus,

c = (2 ± √67)/9
The values of $$ c $$ in $$(-1, 2)$$ that satisfy the Mean Value Theorem are $$ c = \frac{2 + \sqrt{67}}{9} $$ and $$ c = \frac{2 - \sqrt{67}}{9} $$.

Problem 8: Let \( f(x)=3 x^{3}-2 x^{2}+x \) on the interval \( [-1,2] \). . Apply the MVT to find the value(s) of \( c \) in \( (-1,2) \) where \( f^{\prime}(c)=\frac{f(2)-f(-1)}{2-(-1)} \).

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