Concavity and the second derivative
Differentiate the slope function once more and you get $f''$, positive where the slope is growing and negative where it is shrinking. This lesson's curve bends downward on the left half, upward on the right, flipping at one point — the inflection point. What is $f''$ at that flip?
Left of zero the cup reads $\cap$ and the readout stays negative — $f''(-1.00) = -2.000$ — with the short tangent segment sitting above the curve. Right of zero the cup flips to $\cup$ and the readout goes positive. At exactly $x = 0$, the marked inflection, f″(x) reads $0.000$ while the curve is still falling: $f'(0) = -1$. The bending changed; the direction did not.
$f'' = (f')'$ measures how the slope itself is changing. Where $f''(x) > 0$ the slope is increasing, so the curve bends upward like a cup: concave up. Where $f''(x) < 0$ it bends downward like a dome: concave down. A point where the concavity switches is an inflection point — here $f''(x) = 2x$ changes sign only at $x = 0$.
Concavity also gives a faster classifier for lesson 3's candidates, the Second Derivative Test: a valley bottom sits inside an upward cup, a hilltop under a dome. When $f'(c) = 0$,
Check it on this curve: $f''(1) = 2 > 0$ and $x = 1$ was the minimum; $f''(-1) = -2 < 0$ and $x = -1$ was the maximum.
At the critical point $x = -1$ the tangent segment goes flat — $f'(-1) = 0$ — and the f″(x) readout shows $-2.000$. What does the Second Derivative Test conclude?
For $-1 < x < 0$ on this curve, $f'(x) < 0$ and $f''(x) < 0$. What is the curve doing over that stretch?
Transfer: $g(x) = x^4$ has $g'(0) = 0$ and $g''(0) = 0$. What does the Second Derivative Test say about $x = 0$?
$f'$ answers direction, $f''$ answers bending, and the two are independent dials. Maxima and minima live where $f'$ crosses zero; inflection points live where $f''$ does. Two sign charts sketch any smooth curve.