1. **State the problem:** We need to check the validity of the Mean Value Theorem (MVT) for the function $f(x) = x^2 - 3x - 1$ on the interval $\left[-\frac{11}{7}, \frac{13}{7}\right]$. 2. **Recall the Mean Value Theorem:** If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists at least one $c \in (a,b)$ such that $$f'(c) = \frac{f(b) - f(a)}{b - a}.$$ 3. **Check continuity and differentiability:** The function $f(x) = x^2 - 3x - 1$ is a polynomial, so it is continuous and differentiable everywhere, including on $\left[-\frac{11}{7}, \frac{13}{7}\right]$. 4. **Calculate $f(a)$ and $f(b)$:** $$a = -\frac{11}{7}, \quad b = \frac{13}{7}$$ $$f\left(-\frac{11}{7}\right) = \left(-\frac{11}{7}\right)^2 - 3\left(-\frac{11}{7}\right) - 1 = \frac{121}{49} + \frac{33}{7} - 1 = \frac{121}{49} + \frac{231}{49} - \frac{49}{49} = \frac{303}{49}$$ $$f\left(\frac{13}{7}\right) = \left(\frac{13}{7}\right)^2 - 3\left(\frac{13}{7}\right) - 1 = \frac{169}{49} - \frac{39}{7} - 1 = \frac{169}{49} - \frac{273}{49} - \frac{49}{49} = -\frac{153}{49}$$ 5. **Calculate the average rate of change:** $$\frac{f(b) - f(a)}{b - a} = \frac{-\frac{153}{49} - \frac{303}{49}}{\frac{13}{7} - \left(-\frac{11}{7}\right)} = \frac{-\frac{456}{49}}{\frac{24}{7}} = -\frac{456}{49} \times \frac{7}{24} = -\frac{3192}{1176} = -\frac{53}{49}$$ 6. **Find $f'(x)$:** $$f'(x) = 2x - 3$$ 7. **Find $c$ such that $f'(c) = -\frac{53}{49}$:** $$2c - 3 = -\frac{53}{49}$$ $$2c = 3 - \frac{53}{49} = \frac{147}{49} - \frac{53}{49} = \frac{94}{49}$$ $$c = \frac{94}{98} = \frac{47}{49}$$ 8. **Check if $c$ lies in $(a,b)$:** $$-\frac{11}{7} \approx -1.571 < \frac{47}{49} \approx 0.959 < \frac{13}{7} \approx 1.857$$ Since $c$ lies in the interval, the Mean Value Theorem holds for the given function and interval. **Final answer:** The Mean Value Theorem is valid for $f(x) = x^2 - 3x - 1$ on $\left[-\frac{11}{7}, \frac{13}{7}\right]$ with $c = \frac{47}{49}$.