The derivative tells you where a function grows and shrinks, but which way the curve bends is told by the sign of the second derivative f″.
Where f″ is positive the graph is concave up like a bowl (∪); where it is negative it is concave down like a lid (∩).
A point where the bending switches direction is an inflection point, and it needs both f″ = 0 and a change of sign of f″ on either side.
In the picture, moving the point along the curve shows the color and the bending switch together as it passes x = 0.
Reading the direction of bending
seegongsik.com
y = x³ − 3x with the tangent at a moving point; for x < 0 it is concave down (orange), for x > 0 concave up (blue), and the origin is the inflection point
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👀 What to read in the picture
①If f″ > 0 the curve is concave up (∪) and the tangent lies below it
②If f″ < 0 it is concave down (∩) and the tangent lies above it
③For y = x³ − 3x, f″(x) = 6x, so the direction of bending changes at x = 0
④That makes the origin an inflection point
Deciding the direction of concavity
Concavity test
f″(x) > 0 ⇒ concave up, f″(x) < 0 ⇒ concave down
The sign of the second derivative fixes which way the curve bends
Condition for an inflection point
f″(a)=0 and f″ changes sign across x=a ⇒ (a, f(a)) is an inflection point
f″ = 0 is only a necessary condition; a change of sign must also be checked
🪞 The sign of f″ sets the shape
①Where f″(x) > 0 the graph is concave up
②Where f″(x) < 0 the graph is concave down
③An inflection point has f″ = 0 and a change of sign; f″ = 0 alone is not enough
④For example, y = x⁴ has f″ = 0 at x = 0 but is concave up on both sides, so it has no inflection point there
Testing extrema with the second derivative
Second-derivative test
f'(a)=0, f″(a)>0 ⇒ local min / f″(a)<0 ⇒ local max
At a point where the derivative is 0, the sign of f″ tells maximum from minimum at once
Worked examples
Example 1
Find the inflection point of f(x)=x³−3x²+1.
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Find the second derivative.
f'(x)=3x²−6x, f″(x)=6x−6
2
Solve f″(x) = 0, check the change of sign, then find the y-coordinate.
f″(x)=0 ⇒ x=1, f(1)=1−3+1=−1
▸ (1, −1)
The sign of f″ changes from − to + across x = 1, so it really is an inflection point.
Example 2
Use the second-derivative test to find the extrema of f(x)=x³−3x.
1
f'(x) = 3x² − 3 = 0 gives x = ±1, and f″(x) = 6x.
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f″(1) = 6 > 0 means a minimum; f″(−1) = −6 < 0 means a maximum.
local min f(1)=−2, local max f(−1)=2
▸ local max 2, local min −2
Without drawing a sign chart, the sign of f″ alone separates the maximum from the minimum. The curve in the picture is this very function.
Summary
Key points
f″>0 concave up, f″<0 concave down, sign-change point = inflection
Sketch a graph by reading f′ (increasing or decreasing) and f″ (concavity and inflection) together
2021 KICE mock exam Math (Calculus) type, adapted
What is the inflection point of f(x)=x³−6x²+9x+1?
①(1, 5)
②(2, 3)
③(2, 5)
④(3, 1)
⑤No inflection point
▸ ② (2, 3)
1
f'(x) = 3x² − 12x + 9 and f″(x) = 6x − 12.
2
f″(x) = 0 gives x = 2, and f(2) = 8 − 24 + 18 + 1 = 3.
inflection point (2, 3)
🎯 Exam Points
①f″ > 0 means concave up, f″ < 0 concave down
②An inflection point needs f″ = 0 together with a change of sign
③Decide extrema at points with f′ = 0 by the sign of f″
④Sketch the graph from the sign charts of f′ and f″
⑤Give an inflection point as coordinates, including the y-value