x = 0
So there is only 1 stationary point, at x = 0
To find y-coordinate of stationary point, calculate f(0)
y = f(0) = 0²/((0+1)(0−1)) = 0
Stationary point: (0, 0)
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Now we use second derivative to determine nature of stationary point:
relative minimum ---> f''(x) > 0
relative maximum ---> f''(x) < 0
point of inflection ----> f''(x) = 0
f'(x) = −2x / (x²−1)²
f''(x) = (−2 * (x²−1)² − (−2x) * 2(x²−1) * 2x) / (x²−1)⁴
f''(x) = 2(x²−1) (−(x²−1) + 2x * 2x) / (x²−1)⁴
f''(x) = 2(x²−1) (−x² + 1 + 4x²) / (x²−1)⁴
f''(x) = 2 (3x² + 1) / (x²−1)³
f''(0) = 2(1)/(−1)³ = −2 < 0 -----> relative maximum at (0, 0)
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Are you sure you have the right function?
There is only 1 stationary point, and f(−4) = 16/15, NOT 8/9