Curve Sketching

Graphing technology can draw any curve in a second, but it only shows the window you pick, and it can hide a turning point or an asymptote just off the screen. With calculus you can find every important feature of a graph from its equation: where it rises and falls, where it turns, how it bends, and what it does far away. This page puts all the tools from the last few lessons into one organized method, so you can sketch polynomial and simple rational functions accurately by hand.

Key ideas

The curve-sketching checklist

Work through these steps in order. Each one adds information to your sketch.

Domain. Polynomials are defined for all real numbers. For a rational function, exclude the zeros of the denominator.
Intercepts. The $y$ -intercept is $f(0)$ . The $x$ -intercepts are the solutions of $f(x) = 0$ . For a polynomial in factored form, note the order of each zero: the graph crosses the axis at a zero of odd order and touches it at a zero of even order (see polynomials in factored form).
Symmetry. If $f(-x) = f(x)$ , the function is even and its graph is symmetric about the $y$ -axis. If $f(-x) = -f(x)$ , it’s odd and the graph has rotational symmetry about the origin. (See even and odd functions.) Symmetry halves the work: anything you find for $x \gt 0$ has a mirror image.
Asymptotes and end behaviour. For a polynomial, the leading term decides what happens as $x \to \pm\infty$ . For a rational function, find vertical asymptotes (zeros of the denominator that aren’t zeros of the numerator) and the horizontal asymptote (compare the degrees), as on the graphs of rational functions page. Check which side of each vertical asymptote goes up and which goes down.
First derivative. Find $f'(x)$ and the critical points. Make a sign chart to get the intervals of increase and decrease, then classify each critical point with the first derivative test.
Second derivative. Find $f''(x)$ . Make a sign chart to get the intervals of concavity and the points of inflection, as in concavity and the second derivative test.
Summary table. Put the signs of $f'$ and $f''$ for every interval in one table, with the shape of the graph on each piece.
Sketch. Draw the asymptotes as dashed lines, plot the intercepts, extrema, and inflection points, then join them with the shapes from the table.

Put every special x-value on the sign charts

The intervals for your sign charts are split by:

critical points (where $f'(x) = 0$ or $f'$ is undefined),
points where $f''(x) = 0$ or $f''$ is undefined,
$x$ -values that aren’t in the domain (vertical asymptotes or holes).

A sign can change at an asymptote even though nothing “happens” there in the usual sense, so the asymptotes go on the chart too. But they are never extrema or points of inflection, because $f$ doesn’t exist there.

The four shapes

Each interval of the summary table has one of four shapes, from the signs of $f'$ and $f''$ :

	$f'' \gt 0$ (concave up)	$f'' \lt 0$ (concave down)
$f' \gt 0$ (increasing)	rising, more and more steeply	rising, but levelling off
$f' \lt 0$ (decreasing)	falling, but levelling off	falling more and more steeply

These are the same connections as in connecting f, f′, and f″, now applied to an equation instead of a graph.

Checking with technology

After sketching by hand, graph the function in Desmos or on a graphing calculator. Your sketch doesn’t need to be to scale, but every feature should match: the same intercepts, extrema, inflection points, and asymptotes, in the same order.

Worked examples

Example 1: A cubic

Sketch $f(x) = x^3 + 3x^2 - 9x - 27$ .

Solution.

Domain: all real numbers.

Intercepts: $f(0) = -27$ . To find the zeros, factor by grouping:

f(x) = x^2(x + 3) - 9(x + 3) = (x^2 - 9)(x + 3) = (x + 3)^2(x - 3)

The zeros are $x = -3$ (order 2, so the graph touches the axis) and $x = 3$ (order 1, so it crosses).

Symmetry: $f(-x) = -x^3 + 3x^2 + 9x - 27$ , which is neither $f(x)$ nor $-f(x)$ . No symmetry.

End behaviour: the leading term is $x^3$ , so $f(x) \to -\infty$ as $x \to -\infty$ and $f(x) \to \infty$ as $x \to \infty$ .

First derivative:

f'(x) = 3x^2 + 6x - 9 = 3(x + 3)(x - 1)

Critical points: $x = -3$ and $x = 1$ . $f(-3) = 0$ and $f(1) = (4)^2(-2) = -32$ .

Second derivative:

f''(x) = 6x + 6 = 6(x + 1)

$f''(x) = 0$ at $x = -1$ , and $f(-1) = (2)^2(-4) = -16$ .

Summary table:

Interval or point	$x \lt -3$	$-3$	$-3 \lt x \lt -1$	$-1$	$-1 \lt x \lt 1$	$1$	$x \gt 1$
$f'$	$+$	$0$	$-$	$-$	$-$	$0$	$+$
$f''$	$-$	$-$	$-$	$0$	$+$	$+$	$+$
Graph of $f$	rising, concave down	local max $(-3, 0)$	falling, concave down	inflection $(-1, -16)$	falling, concave up	local min $(1, -32)$	rising, concave up

$f'$ changes from $+$ to $-$ at $x = -3$ (local maximum) and from $-$ to $+$ at $x = 1$ (local minimum). $f''$ changes sign at $x = -1$ , so $(-1, -16)$ is a point of inflection.

Sketch: plot the five key points and join them with the shapes in the table.

f(x) = (x + 3)^2(x - 3)

, with every feature found from the equation. The two axes use different scales.

Check: the inflection point of a cubic is always halfway between its turning points, and $-1$ is halfway between $-3$ and $1$ . ✓

Example 2: An even quartic

Sketch $f(x) = x^4 - 8x^2$ .

Solution.

Domain: all real numbers.

Intercepts: $f(0) = 0$ . Zeros: $x^2(x^2 - 8) = 0$ , so $x = 0$ (order 2, touches) and $x = \pm\sqrt{8} = \pm 2\sqrt{2} \approx \pm 2.828$ .

Symmetry: $f(-x) = (-x)^4 - 8(-x)^2 = x^4 - 8x^2 = f(x)$ , so $f$ is even. The graph is symmetric about the $y$ -axis.

End behaviour: the leading term is $x^4$ , so $f(x) \to \infty$ at both ends.

First derivative:

f'(x) = 4x^3 - 16x = 4x(x - 2)(x + 2)

Critical points: $x = -2, 0, 2$ , with $f(0) = 0$ and $f(\pm 2) = 16 - 32 = -16$ .

Interval	$x \lt -2$	$-2 \lt x \lt 0$	$0 \lt x \lt 2$	$x \gt 2$
Sign of $f'$	$-$	$+$	$-$	$+$

So $f$ is decreasing on $(-\infty, -2)$ and $(0, 2)$ , and increasing on $(-2, 0)$ and $(2, \infty)$ . Local minimums at $(-2, -16)$ and $(2, -16)$ ; local maximum at $(0, 0)$ .

Second derivative:

f''(x) = 12x^2 - 16 = 4(3x^2 - 4)

$f''(x) = 0$ when $x^2 = \dfrac{4}{3}$ , so $x = \pm\dfrac{2}{\sqrt{3}} = \pm\dfrac{2\sqrt{3}}{3} \approx \pm 1.155$ . Then $f = \dfrac{16}{9} - \dfrac{32}{3} = -\dfrac{80}{9} \approx -8.889$ .

$f''$ is positive when $\lvert x \rvert \gt 1.155$ and negative between. So $f$ is concave up on $\left(-\infty, -\frac{2\sqrt{3}}{3}\right)$ and $\left(\frac{2\sqrt{3}}{3}, \infty\right)$ , and concave down on $\left(-\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}\right)$ . The points of inflection are $\left(\pm\dfrac{2\sqrt{3}}{3}, -\dfrac{80}{9}\right) \approx (\pm 1.155, -8.889)$ .

Summary table (for $x \ge 0$ ; the left half is its mirror image):

Interval or point	$0$	$0 \lt x \lt 1.155$	$1.155$	$1.155 \lt x \lt 2$	$2$	$x \gt 2$
$f'$	$0$	$-$	$-$	$-$	$0$	$+$
$f''$	$-$	$-$	$0$	$+$	$+$	$+$
Graph of $f$	local max $(0, 0)$	falling, concave down	inflection	falling, concave up	local min $(2, -16)$	rising, concave up

Sketch: a “W” shape: down from the top left to $(-2, -16)$ , up to touch the origin, down to $(2, -16)$ , and up to the top right, crossing the $x$ -axis at $\pm 2.828$ .

Example 3: A rational function

Sketch $f(x) = \dfrac{x^2}{x^2 - 4}$ .

Solution.

Domain: $x^2 - 4 = (x - 2)(x + 2) \ne 0$ , so the domain is $\{x \in \mathbb{R} \mid x \ne \pm 2\}$ .

Intercepts: $f(0) = 0$ , and $f(x) = 0$ only when $x = 0$ . The only intercept is the origin.

Symmetry: $f(-x) = \dfrac{x^2}{x^2 - 4} = f(x)$ , so $f$ is even.

Asymptotes: vertical asymptotes $x = -2$ and $x = 2$ (the numerator isn’t $0$ there). The numerator and denominator both have degree $2$ with leading coefficients $1$ , so the horizontal asymptote is $y = 1$ . Near the vertical asymptote $x = 2$ :

just right of $2$ (try $x = 2.1$ ): $\dfrac{+}{\text{small } +}$ , so $f(x) \to \infty$ ;
just left of $2$ (try $x = 1.9$ ): $\dfrac{+}{\text{small } -}$ , so $f(x) \to -\infty$ .

By symmetry, $f(x) \to \infty$ just left of $-2$ and $f(x) \to -\infty$ just right of $-2$ . Also, $f(x) - 1 = \dfrac{4}{x^2 - 4}$ , which is positive for $\lvert x \rvert \gt 2$ , so the outer branches approach $y = 1$ from above.

First derivative (quotient rule):

f'(x) = \frac{2x(x^2 - 4) - x^2(2x)}{(x^2 - 4)^2} = \frac{-8x}{(x^2 - 4)^2}

The denominator is positive wherever $f$ is defined, so the sign of $f'$ is the sign of $-8x$ : positive for $x \lt 0$ , negative for $x \gt 0$ . The only critical point is $x = 0$ .

$f$ is increasing on $(-\infty, -2)$ and $(-2, 0)$ , and decreasing on $(0, 2)$ and $(2, \infty)$ . Local maximum at $(0, 0)$ .

Second derivative:

f''(x) = \frac{8(3x^2 + 4)}{(x^2 - 4)^3}

The numerator is always positive, so $f''$ has the sign of $(x^2 - 4)^3$ , which is the sign of $x^2 - 4$ . So $f$ is concave up on $(-\infty, -2)$ and $(2, \infty)$ , and concave down on $(-2, 2)$ . The concavity changes only at the asymptotes, so there are no points of inflection.

Summary table:

Interval or point	$x \lt -2$	$-2 \lt x \lt 0$	$0$	$0 \lt x \lt 2$	$x \gt 2$
$f'$	$+$	$+$	$0$	$-$	$-$
$f''$	$+$	$-$	$-$	$-$	$+$
Graph of $f$	rising, concave up	rising, concave down	local max $(0, 0)$	falling, concave down	falling, concave up

f(x) = \dfrac{x^2}{x^2 - 4}

: two vertical asymptotes, a horizontal asymptote

y = 1

, and a local maximum at the origin.

The range is $\{y \in \mathbb{R} \mid y \le 0 \text{ or } y \gt 1\}$ .

Example 4: From derivative information

A polynomial function $f$ has $f'(x) = 4 - x^2$ .

(a) Find the intervals of increase and decrease, the $x$ -values of the local extrema, and the intervals of concavity.
(b) Explain why infinitely many graphs fit this information.
(c) Suppose also that $f(0) = 1$ . Check that $f(x) = 4x - \dfrac{x^3}{3} + 1$ works, and find the coordinates of the key points.

Solution.

(a) $f'(x) = (2 - x)(2 + x)$ , so the critical points are $x = \pm 2$ .

Interval	$x \lt -2$	$-2 \lt x \lt 2$	$x \gt 2$
Sign of $f'$	$-$	$+$	$-$

$f$ is decreasing on $(-\infty, -2)$ and $(2, \infty)$ , and increasing on $(-2, 2)$ . There’s a local minimum at $x = -2$ and a local maximum at $x = 2$ .

$f''(x) = -2x$ , which is positive for $x \lt 0$ and negative for $x \gt 0$ . So $f$ is concave up on $(-\infty, 0)$ and concave down on $(0, \infty)$ , with a point of inflection at $x = 0$ .

(b) The derivative tells you the shape (slopes) of the graph but not its height. Moving the graph up or down by any constant doesn’t change any slope, so every vertical translation of one solution is another solution. For example, a graph with its inflection point at $(0, 0)$ and one with it at $(0, 5)$ both fit.

f(-2) = -8 + \frac{8}{3} + 1 = -\frac{13}{3}, \qquad f(0) = 1, \qquad f(2) = 8 - \frac{8}{3} + 1 = \frac{19}{3}

Local minimum $\left(-2, -\dfrac{13}{3}\right)$ , point of inflection $(0, 1)$ , local maximum $\left(2, \dfrac{19}{3}\right)$ . The graph comes down from the top left (the leading term is $-\frac{x^3}{3}$ ), turns at the minimum, rises through the inflection point, turns at the maximum, and falls to the bottom right.

Common mistakes

Calling an asymptote a critical point or an inflection point. In Example 3, the concavity changes at $x = \pm 2$ , but $f$ isn’t defined there, so there’s no point of inflection. Put asymptotes on your sign charts to split the intervals, but never list them as features of the graph.

Writing intervals across an asymptote. ” $f$ is increasing on $(-\infty, 0)$ ” is wrong in Example 3, because $f$ isn’t defined at $-2$ . Write the two intervals separately: $(-\infty, -2)$ and $(-2, 0)$ .

Finding x-values but not points. A local maximum or a point of inflection is a point on the graph. Substitute into the original $f(x)$ , not into $f'$ or $f''$ , to get the $y$ -coordinate.

Assuming f″ = 0 means an inflection point. Check that $f''$ actually changes sign. For $f(x) = 3x^4 - 4x^3$ (Practice 3), it does at $x = 0$ ; for $f(x) = x^4$ , it doesn’t.

Skipping the shape between points. Plotting the key points and joining them with straight lines loses the curve’s character. Use the summary table: “falling, concave up” means the graph levels off as it approaches a minimum, not that it hits it at a sharp angle.

Sketching only what the calculator window shows. A graphing window can hide an extremum or make an asymptote look like a steep line. Find the features with calculus first, then use technology to check.

Practice

1. (Warm-up) For $f(x) = x^3 - 3x$ , find the intercepts, decide whether $f$ is even, odd, or neither, and describe the end behaviour.

Solution

$f(0) = 0$ . Zeros: $x(x^2 - 3) = 0$ , so $x = 0$ and $x = \pm\sqrt{3} \approx \pm 1.732$ .

$f(-x) = -x^3 + 3x = -(x^3 - 3x) = -f(x)$ , so $f$ is odd: its graph has rotational symmetry about the origin.

The leading term is $x^3$ , so $f(x) \to -\infty$ as $x \to -\infty$ and $f(x) \to \infty$ as $x \to \infty$ .

2. (Warm-up) For $f(x) = 2x^3 + 3x^2 - 12x$ , find the local extrema and the point of inflection.

Solution

$f'(x) = 6x^2 + 6x - 12 = 6(x + 2)(x - 1)$ , so the critical points are $x = -2$ and $x = 1$ .

$f''(x) = 12x + 6$ . Since $f''(-2) = -18 \lt 0$ , there’s a local maximum at $(-2, f(-2)) = (-2, 20)$ . Since $f''(1) = 18 \gt 0$ , there’s a local minimum at $(1, -7)$ .

$f''(x) = 0$ at $x = -\dfrac{1}{2}$ , and $f''$ changes from negative to positive there. $f\left(-\frac{1}{2}\right) = -\frac{1}{4} + \frac{3}{4} + 6 = \frac{13}{2}$ , so the point of inflection is $\left(-\dfrac{1}{2}, \dfrac{13}{2}\right)$ .

3. (Core) Sketch $f(x) = 3x^4 - 4x^3$ . Show the intercepts, intervals of increase and decrease, local extrema, intervals of concavity, and points of inflection.

Solution

Domain: all real numbers. Intercepts: $f(x) = x^3(3x - 4)$ , so $(0, 0)$ (order 3, crosses while flattening) and $\left(\frac{4}{3}, 0\right)$ . Symmetry: none. End behaviour: $f(x) \to \infty$ at both ends.

$f'(x) = 12x^3 - 12x^2 = 12x^2(x - 1)$ . The factor $12x^2$ is never negative, so $f' \lt 0$ for $x \lt 1$ (except $f'(0) = 0$ ) and $f' \gt 0$ for $x \gt 1$ . $f$ is decreasing on $(-\infty, 1)$ and increasing on $(1, \infty)$ . Local minimum at $(1, -1)$ . At $x = 0$ , $f'$ doesn’t change sign, so there’s no extremum there.

$f''(x) = 36x^2 - 24x = 12x(3x - 2)$ : positive for $x \lt 0$ , negative for $0 \lt x \lt \frac{2}{3}$ , positive for $x \gt \frac{2}{3}$ . Concave up on $(-\infty, 0)$ and $\left(\frac{2}{3}, \infty\right)$ , concave down on $\left(0, \frac{2}{3}\right)$ . Points of inflection: $(0, 0)$ and $\left(\dfrac{2}{3}, -\dfrac{16}{27}\right) \approx (0.667, -0.593)$ .

Interval	$x \lt 0$	$0 \lt x \lt \frac{2}{3}$	$\frac{2}{3} \lt x \lt 1$	$x \gt 1$
$f'$	$-$	$-$	$-$	$+$
$f''$	$+$	$-$	$+$	$+$
Shape	falling, concave up	falling, concave down	falling, concave up	rising, concave up

Sketch: coming down from the top left, the graph flattens to a horizontal tangent at the origin (an inflection point), keeps falling more steeply, changes concavity at $\left(\frac{2}{3}, -\frac{16}{27}\right)$ , reaches its minimum at $(1, -1)$ , then rises through $\left(\frac{4}{3}, 0\right)$ .

4. (Core) Sketch $f(x) = -x^3 + 3x + 2$ . (Hint: $f(x) = -(x + 1)^2(x - 2)$ .)

Solution

Intercepts: $y$ -intercept $(0, 2)$ ; zeros $x = -1$ (order 2, touches) and $x = 2$ (crosses). Symmetry: none. End behaviour: leading term $-x^3$ , so $f(x) \to \infty$ as $x \to -\infty$ and $f(x) \to -\infty$ as $x \to \infty$ .

$f'(x) = -3x^2 + 3 = -3(x - 1)(x + 1)$ : negative for $x \lt -1$ , positive on $(-1, 1)$ , negative for $x \gt 1$ . Local minimum $(-1, 0)$ , local maximum $(1, 4)$ .

$f''(x) = -6x$ : concave up on $(-\infty, 0)$ , concave down on $(0, \infty)$ . Point of inflection $(0, 2)$ .

Interval	$x \lt -1$	$-1 \lt x \lt 0$	$0 \lt x \lt 1$	$x \gt 1$
$f'$	$-$	$+$	$+$	$-$
$f''$	$+$	$+$	$-$	$-$
Shape	falling, concave up	rising, concave up	rising, concave down	falling, concave down

Sketch: down from the top left to touch the axis at the minimum $(-1, 0)$ , up through the inflection point $(0, 2)$ to the maximum $(1, 4)$ , then down through $(2, 0)$ .

5. (Core) Sketch $f(x) = \dfrac{8}{x^2 + 4}$ . State its range.

Solution

Domain: $x^2 + 4 \gt 0$ always, so all real numbers, with no vertical asymptotes. Intercepts: $(0, 2)$ ; no $x$ -intercepts, since the numerator is never $0$ . Symmetry: $f(-x) = f(x)$ , so even. Horizontal asymptote: the degree of the denominator is bigger, so $y = 0$ . Since $f(x) \gt 0$ , the graph approaches it from above.

$f'(x) = \dfrac{-16x}{(x^2 + 4)^2}$ : positive for $x \lt 0$ , negative for $x \gt 0$ . Increasing on $(-\infty, 0)$ , decreasing on $(0, \infty)$ , local (and absolute) maximum $(0, 2)$ .

f''(x) = \frac{16(3x^2 - 4)}{(x^2 + 4)^3}

$f''(x) = 0$ at $x = \pm\dfrac{2\sqrt{3}}{3} \approx \pm 1.155$ , where $f = \dfrac{8}{\frac{4}{3} + 4} = \dfrac{3}{2}$ . Concave up for $\lvert x \rvert \gt 1.155$ , concave down between. Points of inflection: $\left(\pm\dfrac{2\sqrt{3}}{3}, \dfrac{3}{2}\right)$ .

Sketch: a bell shape, peaking at $(0, 2)$ , bending from concave down to concave up at height $\frac{3}{2}$ , and levelling off toward the $x$ -axis on both sides.

Range: $\{y \in \mathbb{R} \mid 0 \lt y \le 2\}$ .

6. (Core) A polynomial function $f$ has $f'(x) = (x - 1)^2(x + 2)$ .

(a) Find the intervals of increase and decrease and classify the critical points.
(b) Find $f''(x)$ and the $x$ -values of the points of inflection.
(c) Describe two different graphs that fit this information.

Solution

(a) Critical points: $x = 1$ and $x = -2$ . The factor $(x - 1)^2$ is never negative, so the sign of $f'$ is the sign of $x + 2$ : negative for $x \lt -2$ , positive for $x \gt -2$ (with $f'(1) = 0$ ). $f$ is decreasing on $(-\infty, -2)$ and increasing on $(-2, \infty)$ . There’s a local minimum at $x = -2$ . At $x = 1$ , $f'$ doesn’t change sign, so it’s not an extremum.

(b) Product rule:

f''(x) = 2(x - 1)(x + 2) + (x - 1)^2 = (x - 1)\big(2(x + 2) + (x - 1)\big) = 3(x - 1)(x + 1)

$f''$ is positive for $x \lt -1$ , negative on $(-1, 1)$ , and positive for $x \gt 1$ , so there are points of inflection at $x = -1$ and $x = 1$ . (At $x = 1$ the tangent is horizontal.)

(c) Both graphs fall to a minimum at $x = -2$ , rise while concave up until $x = -1$ , rise while concave down until they flatten to a horizontal tangent at $x = 1$ , then rise more and more steeply. One graph could have its minimum at $(-2, -5)$ and the other at $(-2, 0)$ : any vertical translation fits, because $f'$ doesn’t determine the height.

7. (Core) Sketch $f(x) = \dfrac{2x}{x^2 - 1}$ .

Solution

Domain: $\{x \in \mathbb{R} \mid x \ne \pm 1\}$ . Intercepts: only $(0, 0)$ . Symmetry: $f(-x) = -f(x)$ , so odd. Asymptotes: vertical $x = -1$ and $x = 1$ ; horizontal $y = 0$ (the denominator has the bigger degree).

Near $x = 1$ : just right, $\frac{+}{\text{small } +} \to \infty$ ; just left, $\frac{+}{\text{small } -} \to -\infty$ . By odd symmetry, just right of $-1$ , $f(x) \to \infty$ , and just left of $-1$ , $f(x) \to -\infty$ .

f'(x) = \frac{2(x^2 - 1) - 2x(2x)}{(x^2 - 1)^2} = \frac{-2(x^2 + 1)}{(x^2 - 1)^2}

$f'(x) \lt 0$ everywhere in the domain, so $f$ is decreasing on $(-\infty, -1)$ , $(-1, 1)$ , and $(1, \infty)$ , with no local extrema.

f''(x) = \frac{4x(x^2 + 3)}{(x^2 - 1)^3}

Sign of $f''$ : negative for $x \lt -1$ , positive on $(-1, 0)$ , negative on $(0, 1)$ , positive for $x \gt 1$ . Concave down on $(-\infty, -1)$ and $(0, 1)$ ; concave up on $(-1, 0)$ and $(1, \infty)$ . The only point of inflection is $(0, 0)$ (the other sign changes happen at asymptotes).

Sketch: the left branch falls from just below $y = 0$ down to $-\infty$ at $x = -1$ . The middle branch comes down from $+\infty$ next to $x = -1$ , passes through the origin (inflection point), and falls to $-\infty$ at $x = 1$ . The right branch comes down from $+\infty$ next to $x = 1$ and levels off just above $y = 0$ .

8. (Challenge) Sketch $f(x) = \dfrac{x}{x^2 + 1}$ , and state its range.

Solution

Domain: all real numbers. Intercepts: $(0, 0)$ only. Symmetry: odd. Asymptotes: horizontal $y = 0$ ; no vertical asymptotes.

f'(x) = \frac{(x^2 + 1) - x(2x)}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}

$f' \gt 0$ on $(-1, 1)$ and $f' \lt 0$ on $(-\infty, -1)$ and $(1, \infty)$ . Local minimum $\left(-1, -\dfrac{1}{2}\right)$ , local maximum $\left(1, \dfrac{1}{2}\right)$ .

f''(x) = \frac{2x(x^2 - 3)}{(x^2 + 1)^3}

$f''(x) = 0$ at $x = 0$ and $x = \pm\sqrt{3}$ . Signs: $-$ for $x \lt -\sqrt{3}$ , $+$ on $(-\sqrt{3}, 0)$ , $-$ on $(0, \sqrt{3})$ , $+$ for $x \gt \sqrt{3}$ . Concave down on $(-\infty, -\sqrt{3})$ and $(0, \sqrt{3})$ ; concave up on $(-\sqrt{3}, 0)$ and $(\sqrt{3}, \infty)$ . Points of inflection: $\left(-\sqrt{3}, -\dfrac{\sqrt{3}}{4}\right)$ , $(0, 0)$ , and $\left(\sqrt{3}, \dfrac{\sqrt{3}}{4}\right) \approx (1.732, 0.433)$ .

Sketch: coming in just below the $x$ -axis from the left, the graph falls to its minimum at $\left(-1, -\frac{1}{2}\right)$ , rises through the origin to its maximum at $\left(1, \frac{1}{2}\right)$ , then falls back toward the $x$ -axis, staying above it.

Range: $\left\{y \in \mathbb{R} \;\middle|\; -\dfrac{1}{2} \le y \le \dfrac{1}{2}\right\}$ .

9. (Challenge) Find the cubic function $f(x) = ax^3 + bx^2 + cx + d$ that has a local maximum at $(0, 4)$ and a local minimum at $(2, 0)$ . Then find its point of inflection.

Solution

$f'(x) = 3ax^2 + 2bx + c$ . Use the four facts:

$f(0) = 4$ gives $d = 4$ .
$f'(0) = 0$ gives $c = 0$ .
$f'(2) = 0$ gives $12a + 4b = 0$ , so $b = -3a$ .
$f(2) = 0$ gives $8a + 4b + 4 = 0$ . Substitute $b = -3a$ : $8a - 12a + 4 = 0$ , so $a = 1$ and $b = -3$ .

f(x) = x^3 - 3x^2 + 4

Check: $f'(x) = 3x^2 - 6x = 3x(x - 2)$ , which changes from $+$ to $-$ at $0$ and from $-$ to $+$ at $2$ . ✓ Also $f(2) = 8 - 12 + 4 = 0$ . ✓

$f''(x) = 6x - 6$ changes sign at $x = 1$ , and $f(1) = 1 - 3 + 4 = 2$ . The point of inflection is $(1, 2)$ , halfway between the two turning points.