1. **Problem statement:** Find approximately the value of $\sqrt[5]{31+2}$ using the Mean Value Theorem (M.V.T). 2. **Rewrite the expression:** We want to approximate $\sqrt[5]{33}$. 3. **Choose a function and point:** Let $f(x) = x^{1/5}$. We know $f(32) = 32^{1/5} = 2$ because $2^5 = 32$. 4. **Apply linear approximation:** Using the derivative, $f'(x) = \frac{1}{5} x^{-4/5}$. At $x=32$, $$f'(32) = \frac{1}{5} \times 32^{-4/5} = \frac{1}{5} \times \frac{1}{(32^{4/5})} = \frac{1}{5} \times \frac{1}{(2^4)} = \frac{1}{5} \times \frac{1}{16} = \frac{1}{80}.$$ 5. **Calculate the change in $x$:** $$\Delta x = 33 - 32 = 1.$$ 6. **Approximate $f(33)$:** $$f(33) \approx f(32) + f'(32) \times \Delta x = 2 + \frac{1}{80} \times 1 = 2 + 0.0125 = 2.0125.$$ 7. **Interpretation:** So, $\sqrt[5]{33} \approx 2.0125$ using the Mean Value Theorem and linear approximation. **Final answer:** $$\boxed{2.0125}$$