calculus
consider the function; f(x)=x^4-3x^2-1
a) Find all the point where f'(x)=0
b) Use the second derivative test to classify the stationary points?
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Answers
a) f'(x) = 4x^3 - 6x = 0 => x = 0, x = ±√(3/2) b) The second derivative test states that if the second derivative (f''(x)) has the same sign as f'(x) at a stationary point, the stationary point is a local minimum and if f''(x) has the opposite sign to f'(x), the stationary point is a local maximum. At x = 0, f''(0) = 12 > 0, so f'(0) = 0 and thus, x = 0 is a local minimum. At x = √(3/2), f''(√(3/2)) = -12 < 0, so f'(√(3/2)) = 0, and thus, x = √(3/2) is a local maximum. At x = -√(3/2), f''(-√(3/2)) = -12 < 0, so f'(-√(3/2)) = 0, and thus, x = -√(3/2) is a local maximum.
