For f(x)=x^3+5x+1, find all possible values c in the interval [-1,1] that satisfy the Mean Value Theorem.

Answer 1

f(x) = x³ + 5x + 1 is continuous on [-1,1] and is differentiable on (-1, 1), so the MVT does apply.

Therefore, there exists at least one number, c, in the interval (-1,1) for which f'(c) = [f(1) - f(-1)] / (1-(-1))

So, 3c²+5 = [7 - (-5] / 2

3c² = 1

c² = 1/3

c = ±√3/3

Answer 2

f(- 1) = - 1 - 5 + 1 = - 5; f(1) = 1 + 5 + 1 = 7; f'(x) = 3x² + 5;

f'(c) = 3c² + 5 = [f(1) - f(- 1)]/(1 + 1) = (7 + 5)/2 = 6; 3c² = 6 - 5; 3c² = 1; c = ±√3/3.