A critical point of a function \(f\) is any value \(x = c\) in its domain where either \(f'(c) = 0\) or \(f'(c)\) does not exist. These are the only places where a local maximum or minimum can occur — so finding critical points is the first step in every optimisation problem.
Think of it this way: at a peak or a valley, the tangent line is perfectly horizontal — slope zero. At a sharp corner (like the tip of \(|x|\)), the derivative doesn't exist. Both are critical points. Neither one is guaranteed to be a max or min — that requires further testing.
At the peak of a hill, the ground is momentarily flat — zero slope in every direction. That's a critical point. Walking down the other side, the slope is negative. So the derivative went from positive (climbing) to zero (peak) to negative (descending). The sign change confirms it's a maximum.
Not every critical point is a max or min. Consider \(f(x) = x^3\) at \(x = 0\): \(f'(0) = 0\), so it's a critical point — but the function keeps increasing through zero. It's a saddle point (also called an inflection point with zero slope). Always test after finding critical points.
− → +: local minimum
no sign change: saddle point
Works even when the second derivative doesn't exist.
f″(c) > 0: local minimum (concave up)
f″(c) = 0: inconclusive — use first derivative test
Faster when f″ is easy to compute.
\(x = \pm 1\)
\(f''(1) = 6 > 0\) → min at \(x=1\)
\(f''(-1) = -6 < 0\) → max at \(x=-1\)
\(f''(0) = 0\) → inconclusive
First deriv test: + → + (no change)
→ saddle point at \(x = 0\)
→ critical point at \(x = 0\)
First deriv: − → + → local min
\(x = 0, \pm\sqrt{2}\)
Test each: max at 0, minima at \(\pm\sqrt{2}\)
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