Shape of a Curve
Critical points, increasing and decreasing, concavity, extrema, and the Mean Value Theorem — reading a function's entire personality from f′ and f″.
01Critical points
A critical point of $f$ is an interior point of the domain where $f'(x) = 0$ or $f'(x)$ doesn't exist. These are the only candidates for local maxima and minima (Fermat's theorem) — a smooth hilltop must have a horizontal tangent, and a corner like $|x|$'s can hide a minimum too. Finding them is step one of every problem in this module: differentiate, set equal to zero, solve, and also list points where the derivative is undefined but the function isn't. Caution: a critical point is only a candidate. $f(x) = x^3$ has $f'(0) = 0$ but no max or min there — the curve just pauses and keeps climbing.
02Increasing, decreasing, and the first derivative test
The sign of $f'$ narrates the function's motion: $f' > 0$ on an interval means $f$ is increasing there; $f' < 0$ means decreasing. The critical points chop the number line into intervals; test one point per interval to build a sign chart. Then the First Derivative Test reads results off the chart: at a critical point $c$,
- $f'$ flips $+$ to $-$ → local maximum at $c$
- $f'$ flips $-$ to $+$ → local minimum at $c$
- no sign change → neither (an $x^3$-style shelf)
Example: $f(x) = x^3 - 3x$ has $f' = 3x^2 - 3 = 3(x-1)(x+1)$: positive, negative, positive across $-1$ and $1$ — so a local max at $x = -1$ and a local min at $x = 1$.
03Concavity and the second derivative
The second derivative narrates the bending. $f'' > 0$ means the slope is increasing — the graph is concave up, cupping like a bowl and lying above its tangent lines. $f'' < 0$ means concave down, doming like a hill. An inflection point is where concavity actually switches — found among points where $f'' = 0$ or is undefined, but confirmed only by a genuine sign change ($f(x) = x^4$ has $f''(0) = 0$ with no inflection). Concavity has a real-world voice: if revenue is rising but concave down, growth is slowing — the derivative of the derivative is the "rate of the rate."
04The second derivative test
Concavity gives a slicker way to classify critical points. If $f'(c) = 0$, then:
$$f''(c) > 0 \;\Rightarrow\; \text{local min} \qquad f''(c) < 0 \;\Rightarrow\; \text{local max} \qquad f''(c) = 0 \;\Rightarrow\; \text{no verdict}$$
Mnemonic: concave up holds water (min); concave down sheds it (max). For $f(x) = x^3 - 3x$: $f'' = 6x$, so $f''(-1) = -6 < 0$ (max ✓) and $f''(1) = 6 > 0$ (min ✓) — agreeing with the first derivative test in two lines. When the test is inconclusive ($f'' = 0$), fall back to the sign chart, which never fails.
05Absolute extrema: the closed interval method
Local extrema are hilltops; absolute extrema are the highest and lowest points overall. The Extreme Value Theorem guarantees both exist when $f$ is continuous on a closed interval $[a,b]$ — the payoff for Module 03's care. Finding them is purely mechanical:
- Find all critical points of $f$ inside $(a, b)$.
- Evaluate $f$ at every critical point and at both endpoints.
- Biggest value wins (absolute max), smallest loses (absolute min).
No sign charts, no classification — just a shortlist and a comparison. Forgetting the endpoints is the classic exam giveaway; on $[0, 3]$, the max of $f(x) = x^2$ is at the endpoint $x = 3$, not at any critical point.
06The Mean Value Theorem
MVT: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c$ in $(a,b)$ satisfies
$$f'(c) = \frac{f(b) - f(a)}{b - a}$$
Somewhere, the instantaneous rate equals the average rate. If you drove 120 miles in 2 hours, at some instant your speedometer read exactly 60 — that's MVT, and it's how a court could prove speeding from toll timestamps. Its quieter consequences run the whole subject: $f' = 0$ everywhere forces $f$ constant, and $f' = g'$ everywhere forces $f = g + C$ — the fact that makes the "+C" of Module 10 honest. (Rolle's Theorem is the special case $f(a) = f(b)$, giving $f'(c) = 0$.)
07Curve sketching: the synthesis
Everything above assembles into a portrait-drawing procedure. From $f$: domain, intercepts, asymptotes (Module 02's limits at infinity). From $f'$: critical points, increasing/decreasing intervals, local extrema. From $f''$: concavity and inflection points. Then draw — plotting the special points first and connecting them respecting every sign. A degree-3 polynomial with two critical points and one inflection, like the hero graph, is the canonical exercise. Curve sketching looks old-fashioned in a graphing-calculator world, but it's really a comprehension test: can you predict the picture before the machine draws it?
08Reading f from f′ — the exam favorite
Many exam problems hand you the graph of $f'$ (not $f$!) and ask about $f$. Translate mechanically: where $f'$ is positive (above the axis), $f$ rises; where $f'$ crosses the axis downward, $f$ has a local max; where $f'$ itself is increasing, $f$ is concave up. The single most common error is treating the graph of $f'$ as if it were $f$ — reading its peaks as $f$'s peaks. Slow down, ask "which function am I looking at?", and convert sign information only.
The portrait protocol. $f'$ zeros and breaks → critical points → sign chart → increasing/decreasing → first derivative test. $f''$ → concavity → inflection points → second derivative test shortcut. Closed interval? Critical points + endpoints, evaluate, compare. And MVT lurks behind any "must there be a moment when…" question.
Find all critical points of $f(x) = x^3 - 3x$.
Solution
$f' = 3x^2 - 3 = 3(x-1)(x+1) = 0 \Rightarrow x = \pm 1$.
Find all critical points of $f(x) = x^{2/3}$.
Solution
$f' = \dfrac{2}{3}x^{-1/3}$ — never zero, but undefined at $x = 0$ while $f(0)$ exists. Critical point: $x = 0$.
Find the intervals where $f(x) = x^3 - 3x$ is increasing and decreasing.
Solution
Sign chart of $3(x-1)(x+1)$: $+$ on $(-\infty,-1)$, $-$ on $(-1,1)$, $+$ on $(1,\infty)$.
Increasing on $(-\infty,-1)\cup(1,\infty)$; decreasing on $(-1,1)$.
Use the First Derivative Test to classify the critical points of $f(x) = x^3 - 3x$.
Solution
At $x=-1$: $f'$ flips $+\to-$ → local max, $f(-1) = 2$.
At $x=1$: $f'$ flips $-\to+$ → local min, $f(1) = -2$.
Use the Second Derivative Test on the critical points of $f(x) = x^4 - 8x^2$.
Solution
$f' = 4x^3 - 16x = 4x(x^2-4) = 0 \Rightarrow x = 0, \pm 2$. $f'' = 12x^2 - 16$.
$f''(0) = -16 < 0$ → local max at $0$. $f''(\pm 2) = 32 > 0$ → local mins at $\pm 2$.
Find the intervals of concavity and inflection points of $f(x) = x^3 - 6x^2 + 5$.
Solution
$f'' = 6x - 12 = 0$ at $x = 2$; negative before, positive after.
Concave down on $(-\infty,2)$, up on $(2,\infty)$; inflection at $(2, -11)$.
Show $f(x) = x^4$ has $f''(0) = 0$ but no inflection point at $0$.
Solution
$f'' = 12x^2 \ge 0$ everywhere — no sign change at $0$, so concavity never flips. ($x=0$ is in fact a local min.)
Find the absolute max and min of $f(x) = x^3 - 3x^2 + 1$ on $[-1, 4]$.
Solution
$f' = 3x^2 - 6x = 3x(x-2) = 0 \Rightarrow x = 0, 2$ (both inside).
$f(-1) = -3$, $f(0) = 1$, $f(2) = -3$, $f(4) = 17$.
Abs max $17$ at $x=4$; abs min $-3$ at $x = -1$ and $x = 2$.
Find the absolute extrema of $f(x) = x + \dfrac{4}{x}$ on $[1, 4]$.
Solution
$f' = 1 - \dfrac{4}{x^2} = 0 \Rightarrow x = 2$ (only $+2$ is in the interval).
$f(1) = 5$, $f(2) = 4$, $f(4) = 5$. Min $4$ at $x=2$; max $5$ at both endpoints.
Verify the MVT for $f(x) = x^2$ on $[1, 3]$: find the promised $c$.
Solution
Average rate: $\dfrac{9 - 1}{3 - 1} = 4$. Set $f'(c) = 2c = 4 \Rightarrow c = 2 \in (1,3)$. ✓
A driver passes two toll booths 90 miles apart, 75 minutes apart, on a road with a 65 mph limit. Prove they sped.
Solution
Average speed $= \dfrac{90}{1.25} = 72$ mph. Position is continuous and differentiable, so MVT gives an instant where the speedometer read exactly $72 > 65$. ∎
Why does the MVT fail for $f(x) = |x|$ on $[-1, 1]$? (Average rate is $0$, but $f' $ is never $0$.)
Solution
$f$ is not differentiable at $x = 0$, which is inside the interval — the differentiability hypothesis fails, so MVT makes no promise.
If $f'(x) = 0$ for all $x$ in an interval, what is $f$? What if two functions have the same derivative everywhere?
Solution
$f$ is constant on the interval (an MVT corollary). Equal derivatives → the functions differ by a constant: $f = g + C$.
Sketch-plan $f(x) = \dfrac{x}{x^2 + 1}$: find its critical points, extrema, and end behavior.
Solution
$f' = \dfrac{1 - x^2}{(x^2+1)^2} = 0 \Rightarrow x = \pm 1$: local min $\left(-1,-\tfrac12\right)$, local max $\left(1,\tfrac12\right)$.
$\lim_{x\to\pm\infty} f = 0$: horizontal asymptote $y = 0$ on both ends — an odd, S-shaped bump.
The graph of $f'$ (not $f$) is positive on $(0, 4)$, crosses zero downward at $x = 4$, and is negative after. What happens to $f$ at $x = 4$?
Solution
$f$ increases before $4$ and decreases after → local maximum of $f$ at $x = 4$.
The graph of $f'$ is increasing on $(2, 7)$. What does that say about $f$ there?
Solution
$f'$ increasing means $f'' > 0$: $f$ is concave up on $(2,7)$ — regardless of whether $f$ is rising or falling.
Find where $f(x) = xe^{-x}$ is increasing, and its maximum value.
Solution
$f' = e^{-x}(1 - x) = 0 \Rightarrow x = 1$; positive before, negative after.
Increasing on $(-\infty, 1)$; absolute max $f(1) = \dfrac{1}{e} \approx 0.37$.
For $f(x) = x^4 - 4x^3$, find all local extrema and inflection points.
Solution
$f' = 4x^3 - 12x^2 = 4x^2(x - 3)$: sign $-,-,+$ across $0, 3$ → no extremum at $0$, local min at $x = 3$ ($f(3) = -27$).
$f'' = 12x^2 - 24x = 12x(x-2)$: sign flips at $0$ and $2$ → inflections at $(0,0)$ and $(2,-16)$.
Rolle's Theorem: show $f(x) = x^3 - x$ satisfies the hypotheses on $[-1, 1]$ and find every valid $c$.
Solution
Polynomial → continuous and differentiable; $f(-1) = f(1) = 0$. ✓
$f'(c) = 3c^2 - 1 = 0 \Rightarrow c = \pm\dfrac{1}{\sqrt3}$, both in $(-1,1)$.
Challenge: a continuous function on $[0,10]$ has $f'(x) > 0$ everywhere except $f'(5) = 0$. Does $f$ have a local max, local min, or neither at $x=5$? Where are its absolute extrema?
Solution
No sign change at $5$ → neither (a shelf). $f$ is increasing overall, so abs min at $x=0$ and abs max at $x=10$ — the endpoints.