Second derivative test — concavity, inflection points: Exercises

Definition Toolkit — everything used on this page, in plain words

$f (x)$ = the height of the road at position $x$ .
$f^{'} (x)$ = the slope = how steep the road is. $f^{'} > 0$ uphill, $f^{'} < 0$ downhill, $f^{'} = 0$ momentarily flat.
$f^{''} (x)$ = the derivative of the slope = how the steepness is changing.
A critical point is a place where $f^{'} (x) = 0$ (the road is flat there). Candidates for peaks and valleys live only here.
Concave up ( $\cup$ , a "cup") means the slope $f^{'}$ is increasing, which happens exactly when $f^{''} > 0$ — the road bends upward and could hold water.
Concave down ( $\cap$ , a "cap") means the slope $f^{'}$ is decreasing, i.e. $f^{''} < 0$ — the road bends downward and sheds water.
An inflection point is a continuous spot where concavity switches ( $\cup \leftrightarrow \cap$ ). It requires both $f^{''} = 0$ (or $f^{''}$ undefined) AND a genuine sign flip of $f^{''}$ .

The figure below is our visual dictionary — every later problem points back to it. Read its parts explicitly:

The x-axis (horizontal) is position $x$ ; the y-axis (vertical) is the height $f (x)$ .
The curve is the cubic $f (x) = x^{3} - 3 x$ .
The pink part of the curve (all $x < 0$ ) is concave down — a cap $\cap$ — and it carries the local max (yellow dot) at $x = - 1$ , height $f (- 1) = 2$ .
The blue part (all $x > 0$ ) is concave up — a cup $\cup$ — and it carries the local min (yellow dot) at $x = 1$ , height $f (1) = - 2$ .
The yellow square at $(0, 0)$ is the inflection point: this is where $f^{''} = 6 x$ changes sign from $-$ (cap) to $+$ (cup). Problem L3.1 re-derives this square with a full sign check.

Figure — Second derivative test — concavity, inflection points

Level 1 — Recognition

(Just read the sign of $f^{''}$ and name the concavity. No calculus test yet.)

L1.1 At a point where $f^{''} (x) = 7$ , is the graph concave up or concave down?

Recall Solution L1.1

$f^{''} = 7 > 0$ . WHAT the sign says: the slope $f^{'}$ is increasing. WHAT it looks like: the road bends like a cup $\cup$ . → Concave up.

L1.2 At a point where $f^{''} (x) = - 0.3$ , name the concavity and say whether a flat point there ( $f^{'} = 0$ ) would be a max or a min.

Recall Solution L1.2

$f^{''} = - 0.3 < 0$ → slope decreasing → cap $\cap$ → concave down. A flat point sitting inside a cap sits at the top of a bump ⇒ local maximum.

L1.3 For $f (x) = x^{2}$ , compute $f^{''}$ and state the concavity everywhere.

Recall Solution L1.3

$f^{'} (x) = 2 x$ , so $f^{''} (x) = 2$ . Since $2 > 0$ for every $x$ , the parabola is concave up everywhere — a single unbroken cup. No inflection point exists (the sign of $f^{''}$ never changes).

Level 2 — Application

(Run the full second-derivative test.)

L2.1 Classify the critical points of $f (x) = x^{2} - 4 x + 1$ .

Recall Solution L2.1

Step 1 (find flat spots). $f^{'} (x) = 2 x - 4 = 0 \Rightarrow x = 2$ . Why: critical points are where the road is flat. Step 2 (concavity there). $f^{''} (x) = 2$ . Why: the test needs the bend at $x = 2$ . Step 3 (verdict). $f^{''} (2) = 2 > 0 \Rightarrow$ local minimum at $x = 2$ . Value: $f (2) = 4 - 8 + 1 = - 3$ .

L2.2 Classify the critical points of $f (x) = - x^{2} + 6 x$ .

Recall Solution L2.2

$f^{'} (x) = - 2 x + 6 = 0 \Rightarrow x = 3$ . $f^{''} (x) = - 2 < 0 \Rightarrow$ local maximum at $x = 3$ . Value $f (3) = - 9 + 18 = 9$ .

L2.3 For $f (x) = x^{3} - 3 x^{2} + 4$ , find and classify all critical points.

Recall Solution L2.3

Step 1. $f^{'} (x) = 3 x^{2} - 6 x = 3 x (x - 2) = 0 \Rightarrow x = 0$ or $x = 2$ . Step 2. $f^{''} (x) = 6 x - 6$ . Step 3. $f^{''} (0) = - 6 < 0 \Rightarrow$ local max at $x = 0$ , value $f (0) = 4$ . $f^{''} (2) = 6 > 0 \Rightarrow$ local min at $x = 2$ , value $f (2) = 8 - 12 + 4 = 0$ .

Level 3 — Analysis

(Chase sign-changes, spot the degenerate / inconclusive cases.)

L3.1 Find all inflection points of $f (x) = x^{3} - 3 x$ and confirm each with a sign check.

Recall Solution L3.1

$f^{''} (x) = 6 x$ . Candidate: $6 x = 0 \Rightarrow x = 0$ . Sign check (this is the whole point): for $x < 0$ , $f^{''} = 6 x < 0$ (cap $\cap$ ); for $x > 0$ , $f^{''} > 0$ (cup $\cup$ ). Sign flips $- \to +$ ⇒ genuine inflection at $(0, 0)$ , since $f (0) = 0$ and $f$ is continuous. This is exactly the yellow square in the figure at the top of the page.

L3.2 Does $f (x) = x^{4}$ have an inflection point at $x = 0$ ? Show your reasoning.

Recall Solution L3.2

$f^{''} (x) = 12 x^{2}$ . Candidate: $12 x^{2} = 0 \Rightarrow x = 0$ . But for $x < 0$ , $12 x^{2} > 0$ ; for $x > 0$ , $12 x^{2} > 0$ . No sign change — concave up on both sides. Therefore no inflection point, even though $f^{''} (0) = 0$ .

L3.3 For $f (x) = x^{4} - 6 x^{2}$ , find every inflection point.

Recall Solution L3.3

Step 1. $f^{'} (x) = 4 x^{3} - 12 x$ , $f^{''} (x) = 12 x^{2} - 12 = 12 (x^{2} - 1) = 12 (x - 1) (x + 1)$ . Step 2. Candidates: $f^{''} = 0 \Rightarrow x = \pm 1$ . Step 3 (sign map of $12 (x - 1) (x + 1)$ ):

$x < - 1$ : both factors negative → product $+$ → $\cup$ .
$- 1 < x < 1$ : $(x - 1) < 0, (x + 1) > 0$ → product $-$ → $\cap$ .
$x > 1$ : both positive → $+$ → $\cup$ .

Sign flips at both $x = - 1$ and $x = + 1$ . Values: $f (\pm 1) = 1 - 6 = - 5$ . Inflections at $(- 1, - 5)$ and $(1, - 5)$ .

L3.4 Where is $f (x) = x^{1/3}$ concave up / down, and is there an inflection at $x = 0$ ?

Recall Solution L3.4

$f^{'} (x) = \frac{1}{3} x^{- 2/3}$ , $f^{''} (x) = - \frac{2}{9} x^{- 5/3}$ . $f^{''}$ does not exist at $x = 0$ — but the definition of an inflection allows " $f^{''} (c) = 0$ or $f^{''} (c)$ undefined." Sign check: for $x > 0$ , $x^{- 5/3} > 0$ so $f^{''} < 0$ (cap $\cap$ ); for $x < 0$ , $x^{- 5/3} < 0$ so $f^{''} > 0$ (cup $\cup$ ). Sign flips $+ \to -$ across $0$ , and $f$ is continuous with $f (0) = 0$ . ⇒ inflection at $(0, 0)$ (a vertical-tangent one).

Level 4 — Synthesis

(Combine the second-derivative test with other tools: the first-derivative test, curve sketching, Taylor.)

L4.1 For $f (x) = x^{4} - 4 x^{3}$ , find all critical points, classify each (using the second-derivative test, and the first-derivative test where the second is inconclusive), and list inflection points.

Recall Solution L4.1

Critical points. $f^{'} (x) = 4 x^{3} - 12 x^{2} = 4 x^{2} (x - 3) = 0 \Rightarrow x = 0$ or $x = 3$ . Second derivative. $f^{''} (x) = 12 x^{2} - 24 x = 12 x (x - 2)$ . At $x = 3$ : $f^{''} (3) = 12 \cdot 3 \cdot 1 = 36 > 0 \Rightarrow$ local min, $f (3) = 81 - 108 = - 27$ . At $x = 0$ : $f^{''} (0) = 0 \Rightarrow$ test inconclusive. Fall back to the First derivative test:

just left of $0$ (say $x = - 1$ ): $f^{'} = 4 (1) (- 4) = - 16 < 0$ (downhill),
just right ( $x = 1$ ): $f^{'} = 4 (1) (- 2) = - 8 < 0$ (still downhill).

Slope stays negative $- \to -$ : no max or min at $0$ — it is a saddle/shelf (a flat spot on a still-descending road). Inflection points. $f^{''} = 12 x (x - 2) = 0 \Rightarrow x = 0, 2$ . Sign of $12 x (x - 2)$ :

$x < 0$ : $(-) (-) = +$ → $\cup$ ; $0 < x < 2$ : $(+) (-) = -$ → $\cap$ ; $x > 2$ : $(+) (+) = +$ → $\cup$ .

Flips at both $x = 0$ and $x = 2$ . So $x = 0$ is a critical point that is also an inflection point. Values $f (0) = 0$ , $f (2) = 16 - 32 = - 16$ . Inflections at $(0, 0)$ and $(2, - 16)$ .

L4.2 Use the Taylor series argument to explain why $f^{''} (c) = 0$ makes the second-derivative test fail, and use it to classify the critical point of $f (x) = x^{4}$ at $x = 0$ .

Recall Solution L4.2

Near a critical point $c$ (where $f^{'} (c) = 0$ ), Taylor gives $f (x) = f (c) + \frac{1}{2} f^{''} (c) (x - c)^{2} + \frac{1}{6} f^{'''} (c) (x - c)^{3} + \frac{1}{24} f^{''''} (c) (x - c)^{4} + \dots$ When $f^{''} (c) \neq = 0$ , the term $\frac{1}{2} f^{''} (c) (x - c)^{2}$ dominates and $(x - c)^{2} \geq 0$ forces the sign of $f (x) - f (c)$ to match $f^{''} (c)$ — clean verdict. When $f^{''} (c) = 0$ the quadratic term vanishes, so it can no longer decide — that is exactly why the test is inconclusive; the next non-zero term takes over. For $f (x) = x^{4}$ at $c = 0$ : $f^{''} (0) = f^{'''} (0) = 0$ but $f^{''''} (0) = 24$ . So $f (x) - f (0) \approx \frac{1}{24} (24) x^{4} = x^{4} \geq 0$ . Since $x^{4} \geq 0$ on both sides, $f (x) \geq f (0)$ nearby ⇒ local minimum. (Matches the first-derivative test.)

L4.3 Sketch-classify $f (x) = \frac{1}{1 + x ^{2}}$ : find its maximum and its inflection points. (Prep for Curve sketching.)

Recall Solution L4.3

First derivative. Using the quotient/chain rule, $f^{'} (x) = \frac{- 2 x}{( 1 + x ^{2} ) ^{2}}$ . Set $= 0 \Rightarrow x = 0$ . Second derivative. $f^{''} (x) = \frac{6 x ^{2} - 2}{( 1 + x ^{2} ) ^{3}}$ . Classify $x = 0$ : $f^{''} (0) = \frac{- 2}{1} = - 2 < 0$ . Why this forces a peak: $f^{''} < 0$ means the road is bending downward (a cap $\cap$ ) right at the flat spot $x = 0$ , so $x = 0$ sits at the top of that downward bend — every nearby point is lower. Hence local (and global) maximum, $f (0) = 1$ . Inflections. Numerator $6 x^{2} - 2 = 0 \Rightarrow x^{2} = \frac{1}{3} \Rightarrow x = \pm \frac{1}{3}$ . Denominator $(1 + x^{2})^{3} > 0$ always, so $f^{''}$ has the sign of $6 x^{2} - 2$ :

$∣ x ∣ < \frac{1}{3}$ : numerator $< 0$ → $\cap$ ; $∣ x ∣ > \frac{1}{3}$ : numerator $> 0$ → $\cup$ .

Sign flips at both $\pm \frac{1}{3}$ . Value $f (\pm \frac{1}{3}) = \frac{1}{1 + \frac{1}{3}} = \frac{3}{4}$ . Inflections at $(\pm \frac{1}{3}, \frac{3}{4})$ .

Level 5 — Mastery

(Full-generality reasoning: cover every case, prove a claim, handle a parameter.)

L5.1 Let $f (x) = x^{3} + b x$ for a real parameter $b$ . Determine, for every value of $b$ , how many local extrema $f$ has and whether $x = 0$ is an inflection point.

Recall Solution L5.1

$f^{'} (x) = 3 x^{2} + b$ , $f^{''} (x) = 6 x$ . Critical points solve $3 x^{2} + b = 0 \Rightarrow x^{2} = - \frac{b}{3}$ .

Case $b > 0$ : $x^{2} = - b /3 < 0$ has no real solution → no critical points → no local extrema. ( $f$ is strictly increasing.)
Case $b = 0$ : $x^{2} = 0 \Rightarrow x = 0$ , but $f^{''} (0) = 0$ → test inconclusive. First-derivative test: $f^{'} = 3 x^{2} \geq 0$ everywhere, no sign change ⇒ shelf, no extremum.
Case $b < 0$ : $x = \pm - b /3$ , two critical points. $f^{''} (+ - b /3) > 0$ ⇒ min; $f^{''} (- - b /3) < 0$ ⇒ max. So exactly two extrema (one min, one max).

Inflection at $x = 0$ : $f^{''} = 6 x$ flips $- \to +$ across $0$ for every $b$ , and $f$ is continuous ⇒ $(0, 0)$ is always an inflection point, regardless of $b$ .

L5.2 Prove: if $f^{''}$ is continuous, $f^{''} (c) = 0$ , and $f^{''}$ changes sign across $c$ , then $c$ is an inflection point; but give a concrete function showing that " $f^{''} (c) = 0$ " alone is not sufficient.

Recall Solution L5.2

Proof of sufficiency. "Concave up" $⟺ f^{''} > 0$ ; "concave down" $⟺ f^{''} < 0$ . If $f^{''}$ changes sign at $c$ (say $-$ on the left, $+$ on the right), then $f$ is concave down immediately left of $c$ and concave up immediately right of $c$ . By the definition of an inflection point (a continuous point where concavity changes), $c$ is an inflection point. Continuity of $f$ at $c$ holds because $f^{''}$ (hence $f^{'}$ and $f$ ) is continuous there. $■$ Necessity fails (" $f^{''} (c) = 0$ " alone): take $f (x) = x^{4}$ . Then $f^{''} (0) = 12 \cdot 0^{2} = 0$ yet $f^{''} (x) = 12 x^{2} \geq 0$ never changes sign → concave up on both sides → no inflection. So the condition $f^{''} (c) = 0$ is necessary-ish (a candidate) but not sufficient.

L5.3 Consider $f (x) = x^{5} - 5 x^{3}$ . Find and fully classify every critical point (including any that need the first-derivative test), and list all inflection points with sign checks.

Recall Solution L5.3

First derivative. $f^{'} (x) = 5 x^{4} - 15 x^{2} = 5 x^{2} (x^{2} - 3) = 0 \Rightarrow x = 0, x = \pm 3$ . Second derivative. $f^{''} (x) = 20 x^{3} - 30 x = 10 x (2 x^{2} - 3)$ . At $x = 3$ : $f^{''} (3) = 103 (2 \cdot 3 - 3) = 103 \cdot 3 = 303 > 0 \Rightarrow$ local min. $f (3) = (3)^{5} - 5 (3)^{3} = 93 - 153 = - 63$ . At $x = - 3$ : $f^{''} (- 3) = - 303 < 0 \Rightarrow$ local max, $f (- 3) = 63$ . At $x = 0$ : $f^{''} (0) = 0$ → inconclusive. First-derivative test on $f^{'} = 5 x^{2} (x^{2} - 3)$ : near $0$ , $5 x^{2} > 0$ and $(x^{2} - 3) < 0$ , so $f^{'} < 0$ on both sides ⇒ no sign change ⇒ shelf, not an extremum. Inflection points. $f^{''} = 10 x (2 x^{2} - 3) = 0 \Rightarrow x = 0$ or $x = \pm 3/2$ . Sign of $10 x (2 x^{2} - 3)$ across the real line (roots at $- 1.5 \approx - 1.22, 0, + 1.5 \approx 1.22$ ):

$x < - 1.5$ : $(-) (+) = -$ → $\cap$
$- 1.5 < x < 0$ : $(-) (-) = +$ → $\cup$
$0 < x < 1.5$ : $(+) (-) = -$ → $\cap$
$x > 1.5$ : $(+) (+) = +$ → $\cup$

Sign flips at all three roots ⇒ inflections at $x = 0, \pm 3/2$ . Values: $f (0) = 0$ ; $f (\pm 3/2) = \pm ((3/2)^{5/2} - 5 (3/2)^{3/2}) = \mp \frac{21}{4} \frac{3}{2}$ (numerically $\approx \mp 6.43$ ). Inflections at $(0, 0), (3/2, - 6.43 \dots), (- 3/2, 6.43 \dots)$ .

Recap ladder

Recall What each level trained

L1 Read the sign of $f^{''}$ → concavity. L2 Run the second-derivative test at a critical point. L3 Chase $f^{''}$ sign-changes; handle $f^{''} = 0$ -no-flip and $f^{''}$ -undefined inflections. L4 Combine with first-derivative test, Taylor, and sketching. L5 Prove the sufficiency of a sign change; sweep all parameter/boundary cases.

Connections

Second Derivative Test — Concavity, Inflection Points
First derivative test
Critical points
Taylor series
Curve sketching
Maxima and minima — optimization
Concavity and convex functions