Find the stationary points of f(x) = x^4 - 4x^2 + 2 and determine their type using f''.

Boost your skills in A Level Further Mathematics Core Pure. Study with flashcards and multiple choice questions, each question is followed by hints and explanations. Get prepared for your exam with confidence!

Multiple Choice

Find the stationary points of f(x) = x^4 - 4x^2 + 2 and determine their type using f''.

To locate stationary points, first find where the slope is zero by solving f'(x) = 0, then use the second derivative to classify each point. For f(x) = x^4 - 4x^2 + 2, compute f'(x) = 4x^3 - 8x = 4x(x^2 - 2). Setting this to zero gives x = 0 and x = ±√2 as stationary points. Next, apply the second derivative test with f''(x) = 12x^2 - 8. Evaluate at the critical points: - at x = 0, f''(0) = -8 < 0, so x = 0 is a local maximum. - at x = ±√2, f''(±√2) = 12(2) - 8 = 16 > 0, so these are local minima. The function values are f(0) = 2 and f(±√2) = -2; since f(x) → +∞ as |x| → ∞, the minima at x = ±√2 are global minima, while the maximum at x = 0 is local.

To locate stationary points, first find where the slope is zero by solving f'(x) = 0, then use the second derivative to classify each point.

For f(x) = x^4 - 4x^2 + 2, compute f'(x) = 4x^3 - 8x = 4x(x^2 - 2). Setting this to zero gives x = 0 and x = ±√2 as stationary points.

Next, apply the second derivative test with f''(x) = 12x^2 - 8. Evaluate at the critical points:

  • at x = 0, f''(0) = -8 < 0, so x = 0 is a local maximum.

  • at x = ±√2, f''(±√2) = 12(2) - 8 = 16 > 0, so these are local minima.

The function values are f(0) = 2 and f(±√2) = -2; since f(x) → +∞ as |x| → ∞, the minima at x = ±√2 are global minima, while the maximum at x = 0 is local.

More practice questions