Concavity and Inflection Points

The second derivative \(f''\) measures how the slope is changing — the bending of the graph. Where \(f'' > 0\) the curve bends up like a cup; where \(f'' < 0\) it bends down like a cap; and where the bending flips, you get an inflection point.

By the end you'll be able to read concavity from the sign of \(f''\), locate inflection points by finding where \(f''\) changes sign, and keep the job of \(f'\) (direction) separate from the job of \(f''\) (bending).

Predict: looking at the curve below, where does it switch from bending down to bending up? Drag the probe (or use the slider) and watch the sign of \(f''\) flip at exactly that spot — the inflection point.

This is \(f(x) = x^3 - 3x\), with \(f''(x) = 6x\). The short thick arc riding on the probe hugs the curve's local bending: it cups upward where f″ > 0 and caps downward where f″ < 0 — the same colors as the shaded regions and the readout. Hover or focus the probe and the white dot for exact values.

x = −1.50 · f′(x) = 3.75 (rising) · f″(x) = −9.00 → negative — concave down

f(x) = x³ − 3x — shaded by the sign of f″(x) = 6x; the bending flips at the inflection point
f(x) bending arc, f″ > 0 bending arc, f″ < 0

The second derivative measures concavity — \(f'' > 0\) means concave up, \(f'' < 0\) means concave down — and an inflection point is where concavity changes.

Formal

\(f''\) is the derivative of \(f'\): it tells you whether the slopes are increasing or decreasing. Where \(f''(x) > 0\), slopes are increasing, and the graph is concave up; where \(f''(x) < 0\), slopes are decreasing, and it's concave down. To map a function's concavity: 1. compute \(f''\); 2. find where \(f'' = 0\) or is undefined — these split the domain into intervals; 3. test the sign of \(f''\) on each interval; 4. declare an inflection point at each place the sign actually changes. Don't skip the points where \(f''\) is undefined — a sign change can happen there too.

Visual

Concave up looks like a cup ∪ — the curve lies above its tangent lines, the way \(x^3 - 3x\) does to the right of the inflection point in the demo. Concave down looks like a cap ∩ — the curve lies below its tangent lines. Notice what concavity is not: direction. In the demo at \(x = -1.5\) the curve is rising (\(f' > 0\)) yet concave down (\(f'' < 0\)) — it climbs while bending over. Direction comes from the sign of \(f'\), the tool you built in the previous lesson; bending comes from the sign of \(f''\). The two are independent, and mixing them up is the most common concavity error.

Caution: \(f''(c) = 0\) alone does not make \(c\) an inflection point — \(f''\) must actually change sign across \(c\). Classic counterexample: \(f(x) = x^4\) has \(f''(x) = 12x^2\), so \(f''(0) = 0\), but \(f'' \ge 0\) on both sides — the graph stays concave up and there's no inflection at 0. (Same trap as the previous lesson: \(f'(c) = 0\) doesn't guarantee a max or min without a sign change of \(f'\).)
Worked example

Find the inflection point of \(f(x) = x^3\). Differentiate twice: \(f'(x) = 3x^2\), \(f''(x) = 6x\). Set \(f'' = 0\): \(6x = 0\) at \(x = 0\). Check the sign change: \(f'' < 0\) on \((-\infty, 0)\) (concave down) and \(f'' > 0\) on \((0, \infty)\) (concave up). The sign changes, so \((0, 0)\) is an inflection point — the graph switches from cap to cup as it passes the origin.

Your turn

Find the inflection point of \(f(x) = x^3 - 6x^2 + 5\). Differentiate twice: \(f'(x) = 3x^2 - 12x\), \(f''(x) = 6x - 12\). Set \(f'' = 0\): \(6x - 12 = 0\) at \(x =\) ____. Now check the sign of \(f''\) on each side and name the inflection point.

Reveal the answer

\(f'' = 0\) at \(x = 2\). For \(x < 2\), \(f'' = 6x - 12 < 0\) (concave down); for \(x > 2\), \(f'' > 0\) (concave up). The sign changes, so \(x = 2\) gives an inflection point at \((2, f(2)) = (2, 8 - 24 + 5) = (2, -11)\).

More info — concavity as "above or below the tangent line"

Here's another way to see it, using the tangent lines you've been drawing since the lesson on reading shape from \(f'\). Draw the tangent at any point of a concave-up stretch: the curve immediately lifts off above the line on both sides, because the slope keeps growing past the tangent's fixed slope. On a concave-down stretch the curve sags below its tangents. An inflection point is where the tangent line switches sides — the curve crosses its own tangent there. The OpenStax section in Dive deeper below proves this version formally.

Check your understanding

Question 1 of 4

Find the inflection point of \(f(x) = x^3 + 3x^2\).

Question 2 of 4

For \(g(x) = (x-1)^4\), \(g''(1) = 0\). Does \(g\) have an inflection point at \(x = 1\)?

Question 3 of 4

On some interval, \(f'(x) > 0\) and \(f''(x) < 0\). Which picture matches the graph of \(f\) there?

Question 4 of 4

A function has \(f''(x) = x^2 - 9\). On which interval is \(f\) concave down?

Recap

  • \(f'' > 0\) ⇒ concave up (cup, above its tangents); \(f'' < 0\) ⇒ concave down (cap, below its tangents).
  • An inflection point is where concavity changes — \(f''\) must change sign, not merely hit zero: \(f(x) = x^4\) has \(f''(0) = 0\) but no inflection.
  • Method: compute \(f''\), split the domain where \(f'' = 0\) or is undefined, test the sign on each interval, and declare an inflection at each sign change.
  • Sign of \(f'\) governs direction (increasing/decreasing); sign of \(f''\) governs bending (concavity) — they're independent features.

Dive deeper

Sources

  • Concavity and Inflection Points