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
$$ f'(c) = \displaystyle{ f(0)-f(-1) \over 0 - (-1) } \ \ \ \ \longrightarrow $$
$$ 3c^2+1 = \displaystyle{ ( (0)^3+ (0) ) - ( (-1)^3 + (-1) ) \over 1 } \ \ \ \ \longrightarrow $$
$$ 3c^2+1 = 0 - ( -2 ) \ \ \ \ \longrightarrow $$
$$ 3c^2+1 = 2 \ \ \ \ \longrightarrow $$
$$ 3c^2 = 1 \ \ \ \ \longrightarrow $$
$$ c^2 = \displaystyle{ 1 \over 3 } \ \ \ \ \longrightarrow $$
$$ c = \displaystyle{ -1 \over \sqrt{3} } \approx -0.577 \ \ \ or \ \ \ c = \displaystyle{ 1 \over \sqrt{3} } \approx 0.577 \ (0.577 \ is \ NOT \ in \ the \ interval \ (-1, 0).) \ \ \ \ \longrightarrow $$
$$ c = \displaystyle{ -1 \over \sqrt{3} } $$
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