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...

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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).