Monotonicity, extrema and inflection points

What is it about?

In a curve sketching investigation you use the derivatives to find where the graph of $f$ rises or falls, where it has maxima and minima and where its curvature changes. The simulation works in seven steps:

Find the derivatives: with the power rule $(x^n)' = n \cdot x^{n-1}$ for each term, e.g. $f(x) = x^3 - 3x$ , $f'(x) = 3x^2 - 3$ , $f''(x) = 6x$ .
Symmetry and behaviour for $x \to \pm\infty$ : only even exponents ⇒ symmetric about the $y$ -axis, only odd ⇒ symmetric about the origin. For $x \to \pm\infty$ the term with the highest exponent decides.
Zeros: $f(x) = 0$ , often by factoring out or by substitution.
Necessary condition: at an extremum the tangent is horizontal, so $f'(x_0) = 0$ . The solutions are only candidates.
Sufficient condition: if $f'$ changes sign at $x_0$ from $+$ to $-$ , there is a maximum, from $-$ to $+$ a minimum. Without a sign change it is a terrace point (saddle point). Cross-check with the second-derivative test: $f''(x_0) < 0$ ⇒ maximum, $f''(x_0) > 0$ ⇒ minimum.
Inflection points: $f''(x_0) = 0$ with a sign change of $f''$ (also sufficient: $f'''(x_0) \neq 0$ ). There the graph switches between concave up ( $f'' > 0$ ) and concave down ( $f'' < 0$ ).
Monotonicity and curvature: where $f' > 0$ , $f$ is strictly increasing, where $f' < 0$ , strictly decreasing.

The sign chart sums up step 5. For $f(x) = x^3 - 3x$ with $f'(x) = 3(x + 1)(x - 1)$ :

$x$	$x < -1$	$-1$	$-1 < x < 1$	$1$	$x > 1$
$f'(x)$	$+$	$0$	$-$	$0$	$+$
$G_f$	increases	$H(-1, 2)$	decreases	$T(1, -2)$	increases

So $f$ is strictly increasing on $(-\infty, -1]$ and $[1, \infty)$ and decreasing on $[-1, 1]$ . From $f''(x) = 6x = 0$ we get the inflection point $W(0, 0)$ . (The simulation uses the German labels H for a maximum, T for a minimum, TP for a terrace point and W for an inflection point.)

Careful with the second-derivative test: if $f''(x_0) = 0$ , it tells you nothing. $f(x) = x^4$ still has a minimum at $0$ ( $f'$ changes from $-$ to $+$ ), but no inflection point, because $f''(x) = 12x^2$ does not change sign. $f(x) = x^3$ has a terrace point at $0$ – that is also an inflection point with a horizontal tangent.

In the simulation the graphs of $f'$ and $f''$ share the $x$ -axis below $f$ . The area between graph and $x$ -axis is green above the axis ( $+$ ) and red below ( $-$ ). Dashed lines link the zeros of $f'$ with the extreme points and those of $f''$ with the inflection points. The point $P$ carries the tangent: its slope is exactly the value of $f'$ below.

Try it

f(x) = x³ − 3x: press “Next step” several times (or tap the dots 1–7 on the card) and watch the sign chart fill cell by cell.
Drag $P$ slowly along the graph or press “Move the tangent”. Where is the bar in the picture of $f'$ longest, where does it vanish? At a horizontal tangent $P$ snaps into place. Tangent at x₀ = 2: check $t(x) = 9x - 16$ .
Drag the point on the $y$ -axis up and down. What changes for $f'$ and $f''$ , what for the positions of the extrema and for their $y$ -values?
Change the coefficient of $x$ in x³ + c·x. From which value of $c$ on do the maximum and minimum disappear?
Terrace point x³ and x⁴: compare the sign charts of $f'$ and $f''$ in each case. Why is $x = 0$ an extremum once and not the other time?

Tasks

Task 1: Full investigation

Load task

Investigate $f(x) = -x^3 + 3x^2$ for extreme and inflection points and state the intervals of monotonicity.

Show solution

$f'(x) = -3x^2 + 6x = -3x(x - 2)$ , candidates $0$ and $2$ . Signs of $f'$ : $- \mid + \mid -$ , so $T(0, 0)$ and $H(2, 4)$ ; cross-check $f''(0) = 6 > 0$ , $f''(2) = -6 < 0$ . $f''(x) = -6x + 6 = 0 \Rightarrow x = 1$ with a sign change: $W(1, 2)$ . Strictly decreasing on $(-\infty, 0]$ and $[2, \infty)$ , increasing on $[0, 2]$ . Show in the simulation

Task 2: W shape

Load task

Find the extreme and inflection points of $f(x) = x^4 - 2x^2$ .

Show solution

$f'(x) = 4x^3 - 4x = 4x(x - 1)(x + 1)$ : candidates $-1$ , $0$ , $1$ . Signs $- \mid + \mid - \mid +$ , so $T(-1, -1)$ , $H(0, 0)$ , $T(1, -1)$ . $f''(x) = 12x^2 - 4 = 0 \Rightarrow x = \pm\frac{1}{\sqrt{3}} \approx \pm 0.58$ with a sign change; $f\left(\pm\frac{1}{\sqrt{3}}\right) = \frac{1}{9} - \frac{2}{3} = -\frac{5}{9} \approx -0.56$ . Because of the symmetry about the $y$ -axis it is enough to investigate $x \geq 0$ . Show in the simulation

Task 3: When the second-derivative test is silent

Show that $f'(x_0) = 0$ is not sufficient for an extremum, and that $f''(x_0) = 0$ does not imply “no extremum”.

Show solution

$f(x) = x^3$ : $f'(0) = 0$ , but $f'(x) = 3x^2$ is positive to the left and right of $0$ – no sign change, so a terrace point $(0, 0)$ . Show $f(x) = x^4$ : $f'(0) = 0$ and $f''(0) = 0$ , yet $f'(x) = 4x^3$ changes from $-$ to $+$ : minimum $(0, 0)$ . Show

Task 4: Curve sketching in seven steps

Load task

Investigate $f(x) = -0.25x^4 + x^3$ completely: symmetry, behaviour for $x \to \pm\infty$ , zeros, extreme and inflection points, monotonicity and curvature.

Show solution

Even and odd exponents: no simple symmetry. Leading coefficient $-0.25$ with degree $4$ : $f(x) \to -\infty$ for $x \to \pm\infty$ . Zeros: $x^3(1 - 0.25x) = 0 \Rightarrow x = 0$ (triple) and $x = 4$ . $f'(x) = -x^3 + 3x^2 = x^2(3 - x)$ : no sign change at $0$ (terrace point $(0, 0)$ ), from $+$ to $-$ at $3$ : $H(3, 6.75)$ , cross-check $f''(3) = -9 < 0$ . $f''(x) = -3x^2 + 6x = -3x(x - 2)$ : $W_1(0, 0)$ and $W_2(2, 4)$ . Strictly increasing on $(-\infty, 3]$ , decreasing on $[3, \infty)$ ; concave down on $(-\infty, 0)$ and $(2, \infty)$ , concave up on $(0, 2)$ . Show in the simulation

Task 5: Make the profile yourself

Load task

Make the profile of $f(x) = x^3 - 3x^2 + 2$ , then reveal the results step by step.

Show solution

$f'(x) = 3x^2 - 6x = 3x(x - 2)$ : $H(0, 2)$ because $f''(0) = -6 < 0$ , and $T(2, -2)$ because $f''(2) = 6 > 0$ . $f''(x) = 6x - 6 = 0 \Rightarrow W(1, 0)$ . Zeros: $x = 1$ and from $x^2 - 2x - 2 = 0$ also $x = 1 \pm \sqrt{3}$ , so $x \approx -0.73$ and $x \approx 2.73$ . No symmetry about the origin or the $y$ -axis (but the graph is symmetric about its inflection point). Show in the simulation

Ideas for the classroom

Use: The simulation fits upper secondary calculus, investigating polynomial functions. Steps 2 and 3 (symmetry, behaviour for $x \to \pm\infty$ , zeros with multiplicity) pick up polynomial functions from lower secondary. It suits introducing, practising and revising before exams.
Work it out first, then reveal: with “Show the results” off, students work out each step on paper, go on with “Next step” and check with “Reveal the results”.
Sketching derivatives: switch off “Show the graphs of f′ and f″ below”, let students sketch $f'$ , then show it – the guide lines show whether the zeros are in the right place.
Exploring parameters: when does $x^3 + cx$ have extreme points? (Only for $c < 0$ .) This prepares families of functions.
Common stumbling blocks:
- The necessary condition $f'(x_0) = 0$ is taken as proof of an extremum.
- From $f''(x_0) = 0$ students conclude too quickly “no extremum” or “inflection point”.
- Maxima and minima are given without their $y$ -coordinate.
- A constant term is overlooked when checking symmetry: $x^3 - 3x + 1$ is not symmetric about the origin.
- Monotonicity is read from the sign of $f$ instead of $f'$ .