Second Derivative & Inflection

The first derivative tells you where a function rises and falls, but which way the curve itself bends is the job of the second derivative f″. Wherever f″ is positive the graph is concave up like a cup (∪), and wherever it is negative it is concave down like a cap (∩). The exact spot where this bending flips over is the inflection point, so a genuine one needs f″=0 together with a real sign change from one side to the other. Move the point slowly along the curve and hunt for the inflection point, where the sign of f″ flips and the color and bending direction turn over together.

Reading the Direction of Bending

Point position x1

👀 See It

①If f″>0 the curve is concave up (∪)

②If f″<0 it is concave down (∩)

③The point where concavity flips is the inflection point

④As you move the point, the spot where the color changes is the inflection point

Defining Concavity

Concavity test

f″(x) > 0 ⇒ concave up, f″(x) < 0 ⇒ concave down

The sign of the second derivative decides 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 necessary — you must also confirm the sign change

🪞 The Sign of f″ Sets the Shape

①Where f″(x)>0 the graph is concave up

②Where f″(x)<0 it is concave down

③An inflection point is where f″=0 AND the sign changes — f″=0 alone is not enough

Second-Derivative Test for Extrema

Second-derivative test

f'(a)=0, f″(a)>0 ⇒ local min / f″(a)<0 ⇒ local max

At a critical point, the sign of f″ quickly tells local max from min

Compute It Directly

Example 1

Find the inflection point of f(x)=x³−3x²+1.

Find the second derivative.

f'(x)=3x²−6x, f″(x)=6x−6

Solve f″(x)=0, confirm the sign change, then find the y-coordinate.

f″(x)=0 ⇒ x=1, f(1)=1−3+1=−1

▸ (1, −1)

Across x=1, f″ changes from − to +, so it is indeed an inflection point.

Example 2

Use the second-derivative test to find the extrema of f(x)=x³−3x.

From f'(x)=3x²−3=0 we get x=±1, and f″(x)=6x.

f″(1)=6>0 gives a local min; f″(−1)=−6<0 gives a local max.

local min f(1)=−2, local max f(−1)=2

▸ local max 2, local min −2

Without a sign chart for f′, a single sign of f″ separates max from min.

Wrap-up

Key result

f″>0 concave up, f″<0 concave down, sign-change point = inflection

Sketch a graph by reading f′ (monotonicity) and f″ (concavity·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)

f'(x)=3x²−12x+9, f″(x)=6x−12.

f″(x)=0 ⇒ x=2, f(2)=8−24+18+1=3.

inflection point (2, 3)

🎯 Exam Points

①f″>0 concave up, f″<0 concave down

②An inflection needs f″=0 AND a sign change (both)

③For extrema, set f'=0 then judge by the sign of f″

④Sketch via a sign chart of f′ and f″

⑤Always find the y-value of the inflection point too

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Applications of Derivatives

Equations & Inequalities

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