Curve sketching

A 3D illustration of red blood cells suspended in a dark red bloodstream — Photo by allinonemovie on Pixabay.

A formula may define a graph exactly without making its shape easy to picture. Plotting a collection of points can help, but individual points do not directly reveal where the graph rises, falls, turns, or changes the way it bends.

Derivatives provide this missing structure. The sign of the first derivative tracks whether the function is increasing or decreasing, while the sign of the second describes how its slope is changing. When we combine that information with intercepts and other known points, we can construct a qualitative sketch: not a perfectly scaled drawing, but a graph that tells the right mathematical story.

By the end of this lesson, you should be able to:

Use the sign of the first derivative to identify increasing and decreasing intervals and classify local extrema.
Use the sign of the second derivative to identify concavity and inflection points.
Combine intercepts and derivative information to sketch and interpret the shape of a curve.

Red blood cells carry a protein called hemoglobin, which picks up oxygen in the lungs and releases it where the body needs it. At any moment, some fraction of hemoglobin's oxygen-carrying sites are occupied. We call that fraction the oxygen saturation: a saturation of $$ 0 $$ means that none of the sites are occupied, while a saturation of $$ 1 $$ means that all of them are.

The availability of oxygen can be measured by its partial pressure, denoted by $$ p $$ , in millimeters of mercury (mmHg). We won't need the physiology behind this measurement; for our purposes, a larger value of $$ p $$ means that more oxygen is available to bind to hemoglobin.

One common approximation has hemoglobin half saturated when $$ p $$ is about $$ 27 $$ mmHg. A simplified model for the saturated fraction is

$S=\frac{p^3}{27^3+p^3}\qquad (p\ge0)\text{.}$

This model is called a Hill function. It is useful because its shape resembles measured oxygen-saturation data: it begins near $$ 0 $$ , rises slowly and then more quickly, and eventually levels off near $$ 1 $$ . Its parameters are adjusted to make that resemblance as close as possible. The exponent summarizes the cooperative binding of oxygen, but it is not a literal count of binding steps. For human adult hemoglobin, a fitted exponent is commonly about $$ 2.7 $$ ; we've used $$ 3 $$ because it preserves the important shape while keeping the mathematics manageable. The NCBI Bookshelf chapter Oxygen Transport provides more information about the oxygen dissociation curve and the Hill model.

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The model's predicted saturated fraction as oxygen pressure increases.

We can calculate a few values to get a feel for how the saturation depends on pressure.

Oxygen pressure $$ p $$	Saturated fraction $$ S $$
$$ 0 $$ mmHg
$$ 13.5 $$ mmHg
$$ 27 $$ mmHg
$$ 54 $$ mmHg

Since the saturation curve is nonlinear, equal changes in pressure lead to different changes in saturation. This leads to a natural question: at what point is hemoglobin saturation most sensitive to oxygen pressure?

To keep the shape visible without carrying $$ 27^3 $$ through every calculation, we'll let $$ x=p/27 $$ . The variable $$ x $$ has no units; it records oxygen pressure as a multiple of $$ 27 $$ mmHg. Thus $$ x=1 $$ corresponds to $$ 27 $$ mmHg, and the same saturation model becomes

$S=\frac{x^3}{1+x^3}\qquad (x\ge0)\text{.}$

The derivative then measures the sensitivity of saturation to this rescaled pressure:

$\displaystyle \dv{S}{x}=\)?Use the Quotient Rule.\(\displaystyle \text{.}$

Changing from $$ p $$ to $$ x $$ rescales the horizontal axis by a positive constant, so it doesn't change which physical pressure corresponds to the greatest sensitivity.

For every $$ x>0 $$ , the numerator and denominator of $\displaystyle \dv{S}{x}$ are positive, so the model predicts that saturation always as more oxygen becomes available.

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The sensitivity

\displaystyle \dv{S}{x}

is shown in blue.

According to the graph, $\displaystyle \dv{S}{x}$ appears to be greatest nearest $$ x= $$ . To locate that input exactly, we need to determine where $\displaystyle \dv{S}{x}$ changes from increasing to decreasing.

The second derivative of $$ S $$ is

$\displaystyle \dv[2]{S}{x}=\)\(\displaystyle \text{.}$

For $$ x>0 $$ , the factors $$ 6x $$ and $$ (1+x^3)^3 $$ are positive. The sign of $\displaystyle \dv[2]{S}{x}$ is therefore controlled entirely by $$ 1-2x^3 $$ , which is zero when

$\displaystyle x=\)\(\displaystyle \text{.}$

This input is approximately $$ 0.794 $$ .

We can now compare the saturation curve with its first two derivatives and analyze its shape:

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The saturated fraction,

\mathhlred{S}

, is red; its derivative,

\displaystyle \mathhlblue{\dv{S}{x}}

, is blue; and its second derivative,

\displaystyle \mathhlgreen{\dv[2]{S}{x}}

, is green.

For $\displaystyle 0<x<\frac{1}{\sqrt[3]{2}}$ , the second derivative $\displaystyle \mathhlgreen{\dv[2]{S}{x}}$ is . Consequently, $\displaystyle \mathhlblue{\dv{S}{x}}$ is , while the graph of $\mathhlred{S}$ is .
For $\displaystyle x>\frac{1}{\sqrt[3]{2}}$ , the second derivative $\displaystyle \mathhlgreen{\dv[2]{S}{x}}$ is . Consequently, $\displaystyle \mathhlblue{\dv{S}{x}}$ is , while the graph of $\mathhlred{S}$ is .

The positive-to-negative change in $\displaystyle \mathhlgreen{\dv[2]{S}{x}}$ means that $\displaystyle \mathhlblue{\dv{S}{x}}$ reaches its maximum at $\displaystyle x=\frac{1}{\sqrt[3]{2}}$ . On the graph of $\mathhlred{S}$ , this is the point where the curve changes from bending upward to bending downward. Such a point is called an inflection point.

Substituting $\displaystyle x=\frac{1}{\sqrt[3]{2}}$ into $\displaystyle S=\frac{x^3}{1+x^3}$ , the saturation fraction at this point is

$\displaystyle S=\)\(\displaystyle \text{.}$

Returning to the original oxygen-pressure context, $$ x=p/27 $$ , so

$p=27x=\frac{27}{\sqrt[3]{2}}\approx21.43\text{ mmHg.}$

According to this model, hemoglobin saturation is most sensitive to oxygen pressure at about $$ 21.43 $$ mmHg, when one-third of its oxygen-carrying sites are occupied. Beyond that pressure, saturation keeps increasing, but it increases more slowly as it approaches its upper limit of $$ 1 $$ .

Notice that the two derivatives tell different parts of this story:

Since $\displaystyle \mathhlblue{\dv{S}{x}}$ is positive, the saturation curve rises throughout its physical domain.
Since $\displaystyle \mathhlgreen{\dv[2]{S}{x}}$ changes from positive to negative, the slopes first increase and then decrease, so the saturation curve changes from bending upward to bending downward.

Together with the intercept $$ (0,0) $$ and the tendency toward $$ S=1 $$ in the long run, this information determines the essential shape of the curve and describes its behavior.

Think About It

Let $$ U=1-S $$ represent the fraction of hemoglobin's oxygen-carrying sites that are unsaturated. Using the graphs above, sketch the graphs of $\displaystyle \dv{U}{x}$ and $\displaystyle \dv[2]{U}{x}$ . How is each graph related to the corresponding derivative graph for $$ S $$ ?
For $\displaystyle x>\frac{1}{\sqrt[3]{2}}$ , a student writes, “It is decreasing because it is negative.” What is unclear or incorrect about this explanation? Rewrite it precisely, replacing each use of “it” with $$ S $$ , $\displaystyle \dv{S}{x}$ , or $\displaystyle \dv[2]{S}{x}$ as appropriate.
The constants in $\displaystyle S=\frac{p^3}{27^3+p^3}$ play different roles in shaping the model. What would happen to the graph if $$ 27 $$ were replaced by a different number? What would happen if the exponent $$ 3 $$ were changed instead?

Increasing and decreasing behavior

A function is increasing on an interval when its outputs increase as the input moves from left to right throughout that interval. It is decreasing on an interval when its outputs decrease. The derivative then gives us a local direction of change at every input in that interval:

If $$ f'(x)>0 $$ throughout an interval, then $$ f $$ is increasing (INC) there.
If $$ f'(x)<0 $$ throughout an interval, then $$ f $$ is decreasing (DEC) there.

To find where the graph might turn, we look for inputs where the rising or falling behavior could change. A horizontal locally matching line is one possibility; a corner where the derivative doesn't exist is another.

A critical point of $$ f $$ is an input $$ c $$ in the domain of $$ f $$ where $$ f'(c)=0 $$ or where $$ f'(c) $$ is undefined. The corresponding output $$ f(c) $$ is called a critical value. Critical points are candidates for where the graph changes direction. For the continuous derivative functions we'll usually analyze, critical points divide the domain into intervals on which the derivative keeps one sign.

The First Derivative Test

A function has a local maximum (MAX) at $$ x=c $$ when $$ f(c) $$ is at least as large as nearby function values. It has a local minimum (MIN) there when $$ f(c) $$ is at least as small as nearby values. To distinguish these possibilities, we look at how the graph moves on either side of the point.

If the graph rises as we approach a point from the left and falls as we leave to the right, the point is a local maximum. If it falls into the point and rises away from it, the point is a local minimum. The First Derivative Test expresses this reasoning using the signs of the derivative.

Example

Find and classify the critical points of $$ f(x)=x^4+2x^3 $$ .

Solution

Differentiation and factoring give $$ f'(x)=4x^3+6x^2=2x^2(2x+3) $$ , so the critical points are $$ -3/2 $$ and $$ 0 $$ . The derivative exists everywhere, so there are no other critical points.

Away from zero, the factor $$ 2x^2 $$ is positive, so the sign of $$ f' $$ comes from $$ 2x+3 $$ . This factor is negative to the left of $$ -3/2 $$ and positive to the right. At $$ 0 $$ , the squared factor makes the derivative zero without changing its sign. We can organize these observations in a sign chart:

$ x $	$ -\infty $	$ -\frac32 $		$ 0 $	$ \infty $
$ f(x) $	DEC	MIN	INC		INC
$ f'(x) $	-	0	+	0	+

The derivative changes from negative to positive at

\( x=-3/2 \)

. At

\( x=0 \)

, it is zero but positive on both sides.

At $$ x=-3/2 $$ , the derivative changes from negative to positive, so $$ f $$ has a local minimum. Its critical value is $\displaystyle f(-3/2)=-\frac{27}{16}$ , giving the point $\displaystyle (-3/2,-27/16)$ on the graph.

At $$ x=0 $$ , the derivative is positive on both sides, so $$ f $$ has neither a local maximum nor a local minimum. The graph has a horizontal tangent but continues increasing through $$ (0,0) $$ . To describe how it bends as it passes through this point, we need information from the second derivative.

Concavity and inflection points

The first derivative tells us whether the graph rises or falls. The second derivative tells us how those slopes change as we move from left to right.

When slopes increase, a falling graph becomes less steep, or a rising graph becomes steeper. In either case, its direction turns upward: this is the bending we call concave up (CCU). When slopes decrease, the direction turns downward, and the graph is concave down (CCD). Thus $$ f''>0 $$ throughout an interval tells us the graph is concave up there, while $$ f''<0 $$ tells us it is concave down.

An inflection point, also called a point of inflection (POI), is a point on a graph where its concavity changes. Both terms are commonly used. Values where $$ f''=0 $$ or $$ f'' $$ does not exist are candidates, but the concavity on both sides must be checked. If an inflection point also has a horizontal tangent, we call its input a saddle point (SP). It is a critical point where the graph changes concavity but continues increasing or decreasing, so it is neither a local maximum nor a local minimum.

Example

Continue with $$ f(x)=x^4+2x^3 $$ . Determine the concavity and inflection points.

Solution

Since $$ f''(x)=12x^2+12x=12x(x+1) $$ , the candidates are $$ x=-1 $$ and $$ x=0 $$ . To the left of $$ -1 $$ , both $$ x $$ and $$ x+1 $$ are negative, so their product is positive. Between $$ -1 $$ and $$ 0 $$ , they have opposite signs; to the right of $$ 0 $$ , both are positive. This gives the concavity chart below.

$ x $	$ -\infty $	$ -1 $		$ 0 $	$ \infty $
$ f(x) $	CCU	POI	CCD	SPPOI	CCU
$ f''(x) $	+	0	-	0	+

The second derivative changes sign at both

\( x=-1 \)

and

\( x=0 \)

Thus the graph is concave up on $(-\infty,-1)$ , concave down on $$ (-1,0) $$ , and concave up again on $(0,\infty)$ . Since $$ f(-1)=-1 $$ and $$ f(0)=0 $$ , the inflection points on the graph are $$ (-1,-1) $$ and $$ (0,0) $$ . Only $$ x=0 $$ is also a critical point, so we can now classify it as a saddle point. The First Derivative Test showed that the graph keeps increasing there; the second derivative shows that it also changes concavity.

The Second Derivative Test

At a critical point where $$ f'(c)=0 $$ , the second derivative can sometimes classify the point immediately.

The test works because a horizontal tangent that bends upward sits at the bottom of a local bowl, while one that bends downward sits at the top of a local cap. The First Derivative Test remains available when the Second Derivative Test is inconclusive.

Reading graphs of derivatives

Sometimes a derivative is supplied as a graph rather than a formula. Where the graph of $$ f' $$ lies above the horizontal axis, $$ f $$ increases; where it lies below, $$ f $$ decreases. Crossings of the axis indicate sign changes, while touches may produce horizontal tangents without extrema.

The graph of $$ f'' $$ can be read in the same way to determine concavity. There is also a useful route when only the graph of $$ f' $$ is available: wherever $$ f' $$ itself rises over an interval, the slopes of $$ f $$ increase, so $$ f $$ is concave up. Wherever $$ f' $$ falls, $$ f $$ is concave down. A rising derivative graph can have a horizontal tangent at an isolated input, so we don't need $$ f'' $$ to be strictly positive at every input to recognize this behavior.

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The graph of

\( f' \)

for the running example

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

Extrema depend on the sign of $$ f' $$ ; concavity depends on whether the values of $$ f' $$ are increasing or decreasing. Keeping those two questions separate makes derivative graphs much easier to interpret.

Constructing a qualitative sketch

A useful curve sketch usually combines the following information:

Identify the domain, intercepts, and any known boundary or long-run behavior.
Find critical points and use $$ f' $$ to determine increasing and decreasing intervals.
Find candidates for changes in concavity and use $$ f'' $$ to determine concavity intervals.
Plot the important points and connect them in a way consistent with every sign chart.

A qualitative sketch need not reproduce exact scale. It must place known points correctly and preserve the required rising, falling, and bending behavior.

For the running example, the intercepts come from $$ f(x)=x^3(x+2) $$ , so the graph crosses the horizontal axis at $$ x=-2 $$ and $$ x=0 $$ . It falls to the local minimum $\displaystyle (-3/2,-27/16)$ , rises through the inflection point $$ (-1,-1) $$ , flattens as it passes through the saddle point $$ (0,0) $$ , and then continues rising. Since the leading term is $$ x^4 $$ , both ends rise without bound. These features determine the graph's essential shape.

Combining the first- and second-derivative information gives one complete sign chart:

$ x $	$ -\infty $	$ -\frac32 $		$ -1 $		$ 0 $	$ \infty $
$ f(x) $	DECCCU	MIN	INCCCU	POI	INCCCD	SPPOI	INCCCU
$ f'(x) $	-	0	+	+	+	0	+
$ f''(x) $	+	+	+	0	-	0	+

The function decreases and is concave up before its minimum, then increases through two changes in concavity. At

\( x=0 \)

, the first derivative is zero without changing sign while the second derivative does change sign.

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The completed graph of

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

agrees with both sign charts. The point

\( (0,0) \)

is an inflection point with a horizontal tangent, so it is a saddle point rather than a local extremum.

A function has $f'(x)${first_relation}0$ and $f''(x)${second_relation}0$ throughout an interval. Describe its graph there.

Choose one The graph is increasing and concave up.The graph is decreasing and concave up.The graph is increasing and concave down.The graph is decreasing and concave down.

Solution

The ${first_sign_word} first derivative means that the graph is ${direction}. The ${second_sign_word} second derivative means that its slopes are ${slope_change}, so the graph is ${concavity}.

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This power-function example is ${direction} and ${concavity} on $$ x>0 $$ .
Analyze $$ g(x)=${g} $$ .
1. Find $$ g'(x) $$ and $$ g''(x) $$ .
  
  $$ g'(x)= $$ and $$ g''(x)= $$ .
  
  Solution
  Differentiating twice gives $g'(x)=${sign_factor}(3x-${a})(x-${a})$ and $$ g''(x)=${gpp} $$ .
2. Determine the increasing and decreasing intervals.
  
  Solution
  The function is ${outside_direction} on $(-\infty,${x_first})$ and $(${a},\infty)$ , and ${middle_direction} on $(${x_first},${a})$ .
3. Find and classify the local extrema.
  
  Solution
  At $x=${x_first}$ , the function changes from ${outside_direction} to ${middle_direction}, giving a ${first_extremum} at $(${x_first},${y_first})$ . At $$ x=${a} $$ , it changes from ${middle_direction} to ${outside_direction}, giving a ${second_extremum} at $$ (${a},0) $$ .
4. Determine the concavity and inflection point.
  
  Solution
  The graph is ${left_concavity} on $(-\infty,${x_inflection})$ and ${right_concavity} on $(${x_inflection},\infty)$ . It has an inflection point at $(${x_inflection},${y_inflection})$ .
5. Sketch a graph consistent with all of this information.
  
  Solution
  
  Since $g(x)=${sign_factor}x(x-${a})^2$ , the sketch passes through $$ (0,0) $$ and touches the horizontal axis at the repeated intercept $$ (${a},0) $$ . It also passes through the ${first_extremum} and inflection point found above.
  
  $ x $ $ -\infty $ $ ${x_first} $ $ ${x_inflection} $ $ ${a} $ $ \infty $
  
  $ g(x) $ POI
  
  $ g'(x) $ 0 0
  
  $ g''(x) $ 0
  
  The generated chart combines the direction and concavity information before the completed graph.
  Loading the graph…If it does not appear, check the Desmos API configuration.
  
  The completed graph agrees with the first- and second-derivative analysis.
Consider $k(x)=${3*a}x^{2/3}+2x$ .

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The graph has a cusp at the origin and a smooth turning point to its left.
1. For $x\ne0$ , $$ k'(x)= $$ . Explain why $$ k'(0) $$ does not exist.
  
  Solution
  For $x\ne0$ , the Power Rule gives $k'(x)=${2*a}x^{-1/3}+2$ . The term $x^{-1/3}=1/\sqrt[3]{x}$ is undefined at $$ x=0 $$ . Geometrically, the graph becomes vertical on either side of the cusp, with no single finite slope at the origin.
2. The critical points of $$ k $$ are .
  
  Solution
  Solving $${2*a}x^{-1/3}+2=0$ gives $x=${turning_x}$ . The input $$ x=0 $$ is also critical because it belongs to the domain of $$ k $$ while $$ k'(0) $$ does not exist.
3. Determine where $$ k $$ is increasing and decreasing, and classify both critical points.
  
  Solution
  The derivative is positive on $(-\infty,${turning_x})$ , negative on $(${turning_x},0)$ , and positive on $(0,\infty)$ . Therefore $$ k $$ has a local maximum at $(${turning_x},${turning_y})$ and a local minimum at the cusp $$ (0,0) $$ .
A student argues: “Since $$ p''(4)=0 $$ , the point $$ (4,p(4)) $$ is an inflection point.” What is missing?

Solution
The student must check the sign of $$ p'' $$ on both sides of $$ 4 $$ . The equation $$ p''(4)=0 $$ only makes $$ 4 $$ a candidate. An inflection point occurs only if the concavity actually changes there.
The derivative of a function is $r'(x)=-(x+${a})(x-${b})^2$ .
1. Determine where $$ r $$ is increasing and decreasing.
  
  Solution
  The squared factor does not change sign at $$ ${b} $$ . The derivative is positive for $$ x<${-a} $$ and negative for $$ x>${-a} $$ , except that it equals zero at $$ ${b} $$ . Hence $$ r $$ increases on $(-\infty,${-a})$ and decreases on $$ (${-a},${b}) $$ and $(${b},\infty)$ .
2. Classify the critical points.
  
  Solution
  
  At $$ x=${-a} $$ , the derivative changes from positive to negative, so $$ r $$ has a local maximum. At $$ x=${b} $$ , the derivative is negative on both sides, so the graph has a horizontal tangent but no local extremum. The derivative $$ r' $$ increases toward $$ 0 $$ just before $$ x=${b} $$ and decreases just after it, so $$ r $$ changes from concave up to concave down there. Thus $$ x=${b} $$ is a saddle point.
  
  $ x $ $ -\infty $ $ ${-a} $ $ ${b} $ $ \infty $
  
  $ r(x) $ INC MAX DEC SP DEC
  
  $ r'(x) $ + 0 - 0 -
  
  The derivative changes sign at $$ x=${-a} $$ but not at $$ x=${b} $$ .
3. Explain why solving $$ r'(x)=0 $$ without a sign chart would not answer the classification question.
  
  Solution
  The equation finds both candidates but does not determine whether the original function turns there. Classifying a maximum or minimum depends on the sign of the derivative on both sides, while recognizing a saddle point also requires determining how the derivative itself changes.
The graph of $$ f' $$ is shown below.

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The graph of $$ f' $$ .
1. Determine where $$ f $$ is increasing and decreasing.
  
  Solution
  The graph shows that $$ f $$ is ${before_direction} on $(-\infty,0)$ and $$ (0,${a}) $$ , continuing in the same direction through $$ 0 $$ . It is ${after_direction} on $(${a},\infty)$ .
2. Find and classify the critical points of $$ f $$ .
  
  Solution
  The critical points are $$ 0 $$ and $$ ${a} $$ . At $$ x=0 $$ , the graph of $$ f' $$ touches the axis without changing sign, so $$ f $$ has no local extremum. Because $$ f' $$ changes direction there, $$ f $$ changes concavity, making $$ 0 $$ a saddle point. At $$ x=${a} $$ , $$ f' $$ changes sign, so $$ f $$ has a ${extremum}.
3. Determine the concavity of $$ f $$ and the inputs where its concavity changes.
  
  Solution
  The graph of $$ f' $$ shows that $$ f $$ is ${outer_concavity} on $(-\infty,0)$ , ${middle_concavity} on $(0,${concavity_change})$ , and ${outer_concavity} again on $(${concavity_change},\infty)$ . Its concavity changes at $$ x=0 $$ and $x=${concavity_change}$ .
4. Sketch one possible graph of $$ f $$ if $$ f(0)=1 $$ .
  
  Solution
  
  A correct qualitative sketch passes through $$ (0,1) $$ and continues ${before_direction} through a horizontal saddle point there. It changes concavity again at $x=${concavity_change}$ , reaches a ${extremum} at $$ x=${a} $$ , and then is ${after_direction}. The derivative graph determines this essential shape, while $$ f(0)=1 $$ fixes its vertical position.
  
  $ x $ $ -\infty $ $ 0 $ $ ${concavity_change} $ $ ${a} $ $ \infty $
  
  $ f(x) $ SPPOI POI
  
  $ f'(x) $ 0 0
  
  $ f''(x) $ 0 0
  
  The chart organizes every conclusion drawn from the generated graph of $$ f' $$ .
Suppose a function $$ q $$ has the following sign information.

$ x $ $ -\infty $ $ -1 $ $ 0 $ $ 2 $ $ \infty $

$ q(x) $ POI

$ q'(x) $ 0 0

$ q''(x) $ 0

The signs and resulting behavior change together when the generated function is reflected.
1. Classify the critical points $$ -1 $$ and $$ 2 $$ .
  
  Solution
  The derivative changes sign at both critical points. The graph has a ${first_extremum} at $$ x=-1 $$ and a ${second_extremum} at $$ x=2 $$ .
2. Where does the graph have an inflection point?
  
  Solution
  The second derivative changes sign at $$ x=0 $$ , so the graph has an inflection point there, provided $$ q(0) $$ exists as stated.
3. Describe a sketch consistent with the table.
  
  Solution
  
  The graph ${outer_direction} and is ${left_concavity} until the ${first_extremum} at $$ x=-1 $$ . It then ${middle_direction}, changes from ${left_concavity} to ${right_concavity} at $$ x=0 $$ , and continues to the ${second_extremum} at $$ x=2 $$ . Afterward, it ${outer_direction} and remains ${right_concavity}.
  
  Loading the graph…If it does not appear, check the Desmos API configuration.
  
  This is one graph consistent with the table. A vertical shift would preserve all the same derivative information.
4. Could the table determine the exact values of $$ q(-1) $$ , $$ q(0) $$ , and $$ q(2) $$ ?
  
  Solution
  No. The sign information determines directions and shape, but it supplies no vertical scale or starting value. Many vertically shifted graphs could share the same derivative signs.
For a real number $$ n $$ , define $$ f(x)=x^n $$ on $$ x>0 $$ .
1. Find $$ f'(x) $$ and $$ f''(x) $$ .
  
  $$ f'(x)= $$ and $$ f''(x)= $$ .
  
  Solution
  The Power Rule gives $f'(x)=nx^{n-1}$ and $f''(x)=n(n-1)x^{n-2}$ .
2. For which values of $$ n $$ is $$ f $$ increasing? For which values is it decreasing? What happens at the remaining value?
  
  Solution
  Because $$ x>0 $$ , the factor $x^{n-1}$ is always positive. Thus $$ f' $$ has the same sign as $$ n $$ . The function is increasing when $$ n>0 $$ , decreasing when $$ n<0 $$ , and constant when $$ n=0 $$ .
3. For which values of $$ n $$ is the graph concave up? For which values is it concave down? Describe the boundary cases.
  
  Solution
  The factor $x^{n-2}$ is positive, so the sign of $$ f'' $$ is the sign of $$ n(n-1) $$ . The graph is concave up when $$ n<0 $$ or $$ n>1 $$ , and concave down when $$ 0<n<1 $$ . When $$ n=0 $$ , the graph is the constant function $$ f(x)=1 $$ ; when $$ n=1 $$ , it is the line $$ f(x)=x $$ . In both boundary cases, $$ f''(x)=0 $$ throughout the domain, so the graph is neither strictly concave up nor strictly concave down.

Oxygen pressure $\( p \)$	Saturated fraction $\( S \)$
$\( 0 \)$ mmHg
$\( 13.5 \)$ mmHg
$\( 27 \)$ mmHg
$\( 54 \)$ mmHg

\( x \)	\( -\infty \)	\( -\frac32 \)		\( 0 \)	\( \infty \)
\( f(x) \)	DEC	MIN	INC		INC
\( f'(x) \)	-	0	+	0	+

\( x \)	\( -\infty \)	\( -1 \)		\( 0 \)	\( \infty \)
\( f(x) \)	CCU	POI	CCD	SPPOI	CCU
\( f''(x) \)	+	0	-	0	+

\( x \)	\( -\infty \)	\( -\frac32 \)		\( -1 \)		\( 0 \)	\( \infty \)
\( f(x) \)	DECCCU	MIN	INCCCU	POI	INCCCD	SPPOI	INCCCU
\( f'(x) \)	-	0	+	+	+	0	+
\( f''(x) \)	+	+	+	0	-	0	+

\( x \)	\( ${x_first} \)	\( ${x_inflection} \)	\( ${a} \)
\( g(x) \)		POI
\( g'(x) \)	0		0
\( g''(x) \)		0

\( x \)	\( -\infty \)	\( ${-a} \)		\( ${b} \)	\( \infty \)
\( r(x) \)	INC	MAX	DEC	SP	DEC
\( r'(x) \)	+	0	-	0	-

\( x \)	\( -1 \)	\( 0 \)	\( 2 \)
\( q(x) \)		POI
\( q'(x) \)	0		0
\( q''(x) \)		0