`f(x) = x^3 + 2x, [-1,1]` Determine whether the Mean Value Theorem can be applied to `f` on the closed interval `[a,b]`. If the Mean Value Theorem can be applied, find all values of `c` in the open interval `(a,b)` such that `f'(c) = (f(b) - f(a))/(b - a)`. If the Mean Value Theorem cannot be applied, explain why not.
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Yes, it can. The function is continuous on [-1, 1] and is differentiable on (-1, 1).
Here a=-1 and b=1. f(a) = f(-1) = -3 and f(b) = f(1) = 3. So `(f(b)-f(a))/(b-a) = 6/2 = 3.`
f'(x) = 3x^2 + 2,
f'(x) = 3 for x^2 = 1/3 and `x=+-1/sqrt(3)` . It is approx. `+-0.58` (two values).
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