SOLUTION 8: We are given the function $f(x) = x^3+x$ and the interval $[-1, 0]$. This function is continuous on the closed interval $[-1, 0]$ since it is a polynomial. The derivative of $f$ is $$f'(x) = 3x^2+1$$ We can now see that $f$ is differentiable on the open interval $(-1, 0)$. The assumptions of the Mean Value Theorem have now been met. Let's apply the Mean Value Theorem and find all possible values of $c$ in the open interval $(-1, 0)$. Then $$f'(c) = \displaystyle{ f(0)-f(-1) \over 0 - (-1) } \ \ \ \ \longrightarrow$$ $$3c^2+1 = \displaystyle{ ( (0)^3+ (0) ) - ( (-1)^3 + (-1) ) \over 1 } \ \ \ \ \longrightarrow$$ $$3c^2+1 = 0 - ( -2 ) \ \ \ \ \longrightarrow$$ $$3c^2+1 = 2 \ \ \ \ \longrightarrow$$ $$3c^2 = 1 \ \ \ \ \longrightarrow$$ $$c^2 = \displaystyle{ 1 \over 3 } \ \ \ \ \longrightarrow$$ $$c = \displaystyle{ -1 \over \sqrt{3} } \approx -0.577 \ \ \ or \ \ \ c = \displaystyle{ 1 \over \sqrt{3} } \approx 0.577 \ (0.577 \ is \ NOT \ in \ the \ interval \ (-1, 0).) \ \ \ \ \longrightarrow$$ $$c = \displaystyle{ -1 \over \sqrt{3} }$$