2 Graph Sketching

Unit I1 Real functions and graphs

2 Graph sketching

After working through this section, you should be able to:

(a) determine the x-intercepts and y-intercept of a given function f ;(b) determine the intervals on which a given function f is positive or

negative;(c) determine the intervals on which a given function f is increasing or

decreasing, and any points at which f has a local maximum or localminimum;

(d) describe the asymptotic behaviour (if any) of a given function f ;(e) sketch the graph of a given function.

In this section we consider the problem of sketching the graphs of functionsthat are rather more complicated than the basic functions. We develop theideas introduced in the previous section, and give a general strategy thatwill enable you to sketch many graphs.

The aim of sketching the graph of a function is to provide a visualsummary of the main properties of the function. Consider, for example,the function

f(x) =1

1 − x2.

By our convention, the domain of this function is the set of all realnumbers excluding 1 and −1; it consists of the three intervals (−∞,−1),(−1, 1) and (1,∞).

A sketch of the graph of this function is shown below.

Several key properties of the function f can be seen from this graph.

1. The domain of f consists of the three intervals

(−∞,−1), (−1, 1), (1,∞).

2. The graph is symmetric about the y-axis, so f is even.

3. The graph of f crosses the y-axis when y = f(0) = 1;The graph of f does not cross the x-axis.

18

Section 2 Graph sketching

4. f takes positive values on the interval (−1, 1);f takes negative values on the intervals (−∞,−1) and (1,∞).

5. f is increasing on the intervals (0, 1) and (1,∞);f is decreasing on the intervals (−∞,−1) and (−1, 0);f has a local minimum at 0.

6. As x approaches 1 from the left or −1 from the right, f(x) becomesvery large and positive;as x approaches 1 from the right or −1 from the left, f(x) becomesvery large and negative;as x becomes large and positive or large and negative, f(x) gets closerand closer to 0.

When sketching a graph, we do not aim to achieve the detailed accuracy ofa computer plot, but instead to represent important features such as thoselisted above. But how do we identify these features?

2.1 Determining features of a graphWe discuss how to determine the possible features of a graph in turn.

DomainWhen the domain of a function is not given, we use our convention andtake the domain to be the set of all real numbers for which the given ruleis applicable. So the domain is the set of all real numbers, excluding anynumbers which give an expression which is not defined—for example, azero in the denominator of a rational function, or the square root of anegative number.

‘Symmetry’ featuresIn the audio section, we saw that a graph may possess certain types ofsymmetry:

• For a periodic function, such as a trigonometric function, the graphis unchanged by a translation along the x-axis through the period,p say:

f(x + p) = f(x).

• For an even function, the graph is unchanged by reflection in they-axis:

f(−x) = f(x).

• For an odd function, the graph is unchanged by rotation through anangle π about the origin:

f(−x) = −f(x).

InterceptsAn intercept is a value of x or y at which the graph y = f(x) of afunction f meets the x- or y-axis, respectively. The x-intercepts are thesolutions (if any) of the equation f(x) = 0 and are also known as the zerosof f . The y-intercept is the value f(0), if this exists.

19


It is usually straightforward to find the y-intercept, but harder to find thex-intercepts, since this involves solving the equation f(x) = 0. Thisequation is not always possible to solve algebraically, but it is usuallypossible to obtain estimates for the solutions by finding intervals of thedomain in which the values of the function f change sign.

Intervals on which a function has constant signWe now consider how to determine the intervals on which the values of afunction have constant sign. We make the following definitions.

Definitions Let f be a real function with domain A. Thenf is positive on an interval I in A if f(x) > 0 for all x in I;f is negative on an interval I in A if f(x) < 0 for all x in I;f has a zero at the point a in A if f(a) = 0.

Sometimes we can find the intervals on which a polynomial or rationalfunction has constant sign by constructing a sign table for f(x).

For example, consider the function

f(x) =1

1 − x2.

We can factorise f(x):

f(x) =1

(1 − x)(1 + x),

and construct the following sign table. In the left-hand column of the tableare the factors that appear in f(x). In the top row are the key values of xat which the factors change sign, and the intervals on either side of thesekey values. The signs of the various factors are shown in the table by +, −or 0, from which the sign of f(x) may be deduced.

x (−∞,−1) −1 (−1, 1) 1 (1,∞)

1 − x + + + 0 −1 + x − 0 + + +

f(x) − ∗ + ∗ −

We use the symbol ∗ to indicatea point which is not in thedomain.

We deduce that

f has no zeros;f is positive on the interval (−1, 1);f is negative on the intervals (−∞,−1) and (1,∞).

20


For quadratic functions, if we cannot factorise the function we cansometimes find out whether it always has the same sign by ‘completing the Frame 2 of the audio section sets

this out asa(x − α)2 + β, where

α =−b

2a, β =

4ac − b2

4a.

You may, of course, use eitherformulation as you prefer.

square’. If the quadratic function is ax2 + bx + c, we can rewrite it asfollows.

ax2 + bx + c = a

(x2 +

b

ax

)+ c = a

(x +

b

2a

)2

− a

(b

2a

)2

+ c

So, for example,

2x2 + 12x + 19 = 2(x2 + 6x) + 19= 2(x + 3)2 − 2 × 32 + 19= 2(x + 3)2 + 1.

From this, we can see that, whatever the value of x,

2x2 + 12x + 19 = 2(x + 3)2 + 1

is always positive.

Exercise 2.1 Complete the square on the following quadraticfunctions.

(a) x2 − 6x + 11 (b) 3x2 + 12x − 1

Intervals on which a function is increasing or decreasingWe referred to functions which are increasing or decreasing on a particularinterval in Section 1, but did not give a definition. We now do this.

Definitions

A function f is increasing on an interval I, if for all x1, x2 ∈ I,

if x1 < x2, then f(x1) ≤ f(x2).

A function f is strictly increasing on an interval I, if for allx1, x2 ∈ I,

if x1 < x2, then f(x1) < f(x2).

A function f is decreasing on an interval I, if for all x1, x2 ∈ I,

if x1 < x2, then f(x1) ≥ f(x2).

A function f is strictly decreasing on an interval I, if for allx1, x2 ∈ I,

if x1 < x2, then f(x1) > f(x2).

We can sometimes determine the intervals on which a function is increasing What it means to bedifferentiable will be defined inthe analysis blocks. For themoment you can assume that afunction is differentiable if it canbe differentiated by the usualmethods.

or decreasing by inspection of the rule of the function. For example, thefunction f(x) = x3 is increasing on R because x3 increases as x increases.

For a differentiable function, however, we can use the derivative of thefunction to identify these intervals.

Increasing/decreasing criterion

1. If f ′(x) > 0 for all x in an interval I, then f is (strictly) increasingon I.

2. If f ′(x) < 0 for all x in an interval I, then f is (strictly)decreasing on I.

21


f is (strictly) increasing on theintervals (a, b) and (c, d).f is (strictly) decreasing on theinterval (b, c).f ′(b) = f ′(c) = 0.

We can determine the intervals on which f is increasing or decreasing bydrawing up a sign table for f ′(x). The table should also include points a atwhich f ′(a) = 0. Such a point is called a stationary point of f ; it is avalue a such that the tangent to the graph is horizontal at the point(a, f(a)).

A stationary point need not be a local maximum or a local minimum. Forexample, f(x) = x3 has a stationary point at 0, with f(0) = 0, but hasneither a local maximum nor a local minimum at 0. In fact, it has what wecall a horizontal point of inflection.

f has a local maximum at b.f has a local minimum at c.f has a horizontal point ofinflection at d.

We can check whether a stationary point is a local maximum, a localminimum or a horizontal point of inflection by using the following test.

First Derivative Test Suppose that a is a stationary point of adifferentiable function f ; that is, f ′(a) = 0.

• If f ′(x) changes from positive to negative as x increasesthrough a, then f has a local maximum at a.

• If f ′(x) changes from negative to positive as x increasesthrough a, then f has a local minimum at a.

• If f ′(x) remains positive or remains negative as x increasesthrough a (except at a itself, where f ′(a) = 0), then f has ahorizontal point of inflection at a. f ′(x) may do none of these

things.

Let us return to the function f(x) =1

1 − x2. To find the intervals on which

the function is increasing and decreasing we use the quotient rule to give You will find a table of standardderivatives in the Handbook,and also a list of the rules fordifferentiating functions. Weassume that you have metdifferentiation in your previousmathematical studies.

f ′(x) = − −2x(1 − x2)2

=2x

(1 − x2)2.

22


The sign table for f ′(x) is as follows.

x (−∞,−1) −1 (−1, 0) 0 (0, 1) 1 (1,∞)

2x − − − 0 + + +(1 − x2)2 + 0 + + + 0 +

f ′(x) − ∗ − 0 + ∗ +

In this case there is no need tofactorise (1 − x2)2, since

(1 − x2)2 ≥ 0.

We find that

f has a stationary point at 0;f is increasing on the intervals (0, 1) and (1,∞);f is decreasing on the intervals (−∞,−1) and (−1, 0).

We deduce that f has a local minimum at 0, by the First Derivative Test.

Asymptotic behaviour of functionsFor a function f , the term asymptotic behaviour refers to the behaviourof points on the graph of y = f(x) for which the variable x or the variabley take arbitrarily large values.

For example, we consider how to determine the features of the graph of the

function f(x) =1

1 − x2as x or y approaches ∞.

The computer plot shown on the left below uses a ‘join-the-dots’ approachto generate the graph of this function. The computer plot is inaccuratenear the ‘missing’ points x = 1 and x = −1, since it shows vertical lines atx = 1 and x = −1 as parts of the graph, whereas we know that thefunction is not defined at these points. It is common for computer plots of A computer plot always assumes

that the curve has no breaks,and tries to join the curve upaccordingly.

graphs to give misleading results near such ‘difficult’ points. By contrast,the sketch of this graph on the right indicates the behaviour of thefunction f near the points 1 and −1 by the use of broken vertical lines.

The plot on the left wasgenerated by Mathcad. We haveadded axes for clarity.

A broken line is used when the graph of a function has anasymptote—that is, a straight line which is approached more and moreclosely by the graph when the domain variable x or the codomain variabley (or both) takes very large values.

An asymptote with an equation of the form x = a is a verticalasymptote. For example, in the above graph, the lines x = −1 and x = 1are vertical asymptotes.

An asymptote with an equation of the form y = b is a horizontalasymptote. For example, in the above graph, the line y = 0 is a When one of the axes is an

asymptote, as in this case, it isnot represented by a broken line.

horizontal asymptote.

23


The behaviour of a function f near a vertical asymptote x = a may takevarious forms. For the above example, we describe the behaviour near thevertical asymptote x = −1 as follows.

f(x) takes arbitrarily large positive valuesas x tends to −1 from the right;

we write this in symbols as

f(x) → ∞ as x → −1+,

and read it as

f(x) tends to infinity as x tends to −1 from the right.

Similarly,

f(x) takes arbitrarily large negative valuesas x tends to −1 from the left;

we write this in symbols as

f(x) → −∞ as x → −1−,

and read it as

f(x) tends to minus infinity as x tends to −1 from the left.

General versions of these statements are illustrated below, together with Intuitive statements of thisnature will be formally definedin Analysis Block A.

similar statements for vertical and for horizontal asymptotes.

In the second example, thegraph crosses the asymptote.

These diagrams illustrate asymptotic behaviour.

Exercise 2.2 Write down four more statements describing the types

of asymptotic behaviour displayed by the function f(x) =1

1 − x2.

There are other types of behaviour that a function may exhibit as thedomain variable x takes large positive or negative values.

24


Exercise 2.3 Describe the asymptotic behaviour of the followingfunctions.

In general, the behaviour of a polynomial function of degree n,

f(x) = anxn + an−1xn−1 + · · · + a1x + a0, where an �= 0,

for large values of x, is similar to that of the term anxn. We call xn the It is xn rather than anxn thatwe are calling ‘the dominantterm’. Thus, for example, theexpressions 3x3 − 2x2 + 1 and−4x3 + 5x − 3 have the samedominant term, namely x3.

dominant term. This behaviour is summarised in the following tables.

an > 0 x → ∞ x → −∞n even f(x) → ∞ f(x) → ∞n odd f(x) → ∞ f(x) → −∞

an < 0 x → ∞ x → −∞n even f(x) → −∞ f(x) → −∞n odd f(x) → −∞ f(x) → ∞

A rational function is a function defined by a rule of the form The function x �−→ 11 − x2

is an

example of a rational function,with p(x) = 1 and q(x) = 1 − x2.x �−→ p(x)

q(x),

where both p and q are polynomial functions. Locating vertical andhorizontal asymptotes is an important step in sketching the graph of anyrational function. Vertical asymptotes occur at the values of x for whichq(x) = 0 and p(x) �= 0, and horizontal asymptotes may occur when x → ∞or x → −∞.

To find the behaviour of a rational function for large values of x, we divideboth the numerator and denominator by the dominant term of thedenominator. If the dominant term of the numerator is a higher powerthan that of the denominator, then there will be no horizontal asymptote,whereas if it is a lower power than that of the denominator, then the liney = 0 will be the horizontal asymptote. The situation is more complicatedif the numerator and denominator have the same dominant term; this isillustrated by later examples.

25


2.2 Strategy for graph sketchingWe begin this subsection by summarising basic features which a sketch of agraph should convey, in the form of a strategy, and then we illustrate thestrategy with several worked examples.

Strategy 2.1 Graph-sketching strategy

To sketch the graph of a given function f , determine the followingfeatures of f (where possible), and show these features in your sketch.

1. The domain of f . All the periodic functions thatyou will meet in this course willinvolve a trigonometric function(sine, cosine, tangent, cotangent,secant or cosecant), so there isno need to investigate whether fis periodic unless one of theseexpressions appears in the rule.

2. Whether f is even, odd or periodic (or none of these).

3. The x-intercepts and y-intercept of f .

4. The intervals on which f is positive or negative.

5. The intervals on which f is increasing or decreasing, the nature ofany stationary points, and the value of f at each of these points.

6. The asymptotic behaviour of f .

Remarks

1. It is important to begin by determining the domain of f . For example,if the domain is [3, 9], then f is neither even nor odd, and you cannotfind the behaviour of f as x → ∞.

2. We have numbered the features for easy reference, but it is notnecessary to find them in the order given above. Indeed, for somegraphs, not all the above features are relevant.

For some graphs, we can obtain sufficient information withoutincluding all the steps. However, it is useful to obtain information inmore than one way, in order to provide a check.

3. Choose the scales on your axes with care: usually, the scales should bethe same on both axes, but it may be necessary to have unequal scalesin order to display some key features of the graph—for example, whenf(x) is much larger than x.

Our first example to illustrate Strategy 2.1 is a polynomial function.

Example 2.1 Sketch the graph of the function

f(x) = 4x3 + 3x2 − 6x + 4.

Solution

1. By our convention, the domain of f is R.

2. The function is neither even nor odd, since, for example,

f(1) = 5, but f(−1) = 9.

There is no trigonometric function involved, so it is not periodic. In future, we shall not normallymention that f is not periodicunless a trigonometric functionis involved.

3. To find the x-intercepts of f we need to solve the equation f(x) = 0,that is

4x3 + 3x2 − 6x + 4 = 0.

There are no obvious factors for the expression on the left, so we We tried various values of x forwhich f(x) was easy to calculate,and chose these particular valuesof x because the values of f arepositive at one value of x andnegative at the other.

cannot easily find the zeros. However,

f(−2) = −4 and f(0) = 4,

26


so f(x) changes from negative to positive as x increases from −2 to 0.Thus there is an x-intercept in the interval (−2, 0). The y-intercept isf(0) = 4.

4. Because we cannot find the zeros of f , we cannot find the intervals onwhich f is positive or negative. However, we know from step 3 that thesign of f changes from negative to positive as x increases from −2 to 0.

5. Differentiating gives

f ′(x) = 12x2 + 6x − 6 = 6(2x2 + x − 1) = 6(2x − 1)(x + 1).

We construct a sign table for f ′(x).

x (−∞,−1) −1 (−1, 12) 1

2 (12 ,∞)

6(2x − 1) − − − 0 +x + 1 − 0 + + +

f ′(x) + 0 − 0 +

From the sign table we see that

f is increasing on the intervals (−∞,−1) and (12 ,∞);

f is decreasing on the interval (−1, 12);

f has stationary points at −1 and 12 .

By the First Derivative Test, we deduce that

there is a local maximum at x = −1 with f(−1) = 9;there is a local minimum at x = 1

2 with f(

12

)= 9

4 .

6. The degree of the polynomial function is odd and the coefficient of x3

is positive, so the first table for the asymptotic behaviour ofpolynomial functions on page 25 gives

f(x) → ∞ as x → ∞ and f(x) → −∞ as x → −∞.

This information enables us to sketch the graph.

The results of steps 4 and 5 show that the graph of f crosses the x-axis atonly one point.

Exercise 2.4 Sketch the graph of the polynomial function Hint: Putting t = x2, theexpression becomes t2 − 2t + 3.

f(x) = x4 − 2x2 + 3.

Next, we sketch the graph of a linear rational function.

27



f(x) =2x − 3x − 1

.

Solution

1. By our convention, the domain of f is R, excluding 1; that is, itconsists of the intervals (−∞, 1) and (1,∞).

2. The function is neither even nor odd, since its domain is notsymmetric about 0; for example,

f(−1) = 52 , but f is not defined at x = 1.

3. f(x) = 0 when 2x − 3 = 0, so the x-intercept is 32 .

f(0) = −3/(−1) = 3, so the y-intercept is 3.

4. We construct a sign table for f(x).

x (−∞, 1) 1 (1, 32) 3

2 (32 ,∞)

2x − 3 − − − 0 +x − 1 − 0 + + +

f(x) + ∗ − 0 +

So

f is positive on the intervals (−∞, 1) and (32 ,∞);

f is negative on the interval (1, 32).

5. Using the rule for differentiating a quotient, we obtain

f ′(x) =(x − 1)2 − (2x − 3)1

(x − 1)2=

1(x − 1)2

.

The derivative f ′ is undefined at 1, and f ′(x) > 0 for x < 1 and x > 1.Thus

f is increasing on the intervals (−∞, 1) and (1,∞);f has no stationary points.

6. The denominator is 0 when x = 1, so

the line x = 1 is a vertical asymptote.

Thus, by the results of step 4 (or step 5), From step 4, f(x) is positive asx tends to 1−, and f(x) isnegative as x tends to 1+.f(x) → ∞ as x → 1− and f(x) → −∞ as x → 1+.

To find the behaviour of f(x) for large positive or negative values of x,we divide both the numerator and denominator of f(x) by thedominant term of the denominator, x: This division is permitted, since

x �= 0.f(x) =

2x − 3x − 1

=2 − 3/x1 − 1/x

.

Now 1/x → 0 as x → ±∞, so We write ‘as x → ±∞’ asshorthand for ‘as x → ∞ andas x → −∞’.f(x) → 2 − (3 × 0)

1 − 0=

2 − 01 − 0

= 2 as x → ±∞.

Thus

the line y = 2 is a horizontal asymptote,

and

f(x) → 2 as x → ±∞.

28



This function can also be written as f(x) = 2 − 1x − 1

.

Exercise 2.5 Sketch the graph of the linear rational function

f(x) =x − 32 − x

.

Exercise 2.6 Sketch the graph of the linear rational function

f(x) =4x + 13x − 5

.

Next, we sketch the graph of a more complicated rational function.


f(x) =x2 − 5x + 4x2 + 5x + 4

.

Solution

1. We factorise f(x) as follows.

f(x) =x2 − 5x + 4x2 + 5x + 4

=(x − 1)(x − 4)(x + 1)(x + 4)

Thus the domain of f is R, excluding −1 and −4; it consists of theintervals (−∞,−4), (−4,−1) and (−1,∞).

2. The function is neither even nor odd, since for example, Alternatively, observe that thedomain is not symmetricabout 0.f(2) = −1

9 , but f(−2) = −9.

3. f(x) = 0 when (x − 1)(x − 4) = 0, so the x-intercepts are 1 and 4.f(0) = 4/4 = 1, so the y-intercept is 1.

4. We construct a sign table for f(x).

x (−∞,−4) −4 (−4,−1) −1 (−1, 1) 1 (1, 4) 4 (4,∞)

x − 1 − − − − − 0 + + +x − 4 − − − − − − − 0 +x + 1 − − − 0 + + + + +x + 4 − 0 + + + + + + +

f(x) + ∗ − ∗ + 0 − 0 +

So

f is positive on the intervals (−∞,−4), (−1, 1) and (4,∞);f is negative on the intervals (−4,−1) and (1, 4).

29


5. Using the quotient rule, we differentiate f(x) as follows.

f ′(x) =(2x − 5)(x2 + 5x + 4) − (x2 − 5x + 4)(2x + 5)

(x2 + 5x + 4)2

=(2x3 + 10x2 + 8x − 5x2 − 25x − 20) − (2x3 − 10x2 + 8x + 5x2 − 25x + 20)

(x2 + 5x + 4)2

=10(x2 − 4)

(x2 + 5x + 4)2

=10(x − 2)(x + 2)(x + 1)2(x + 4)2

We construct a sign table for f ′(x).

x (−∞,−4) −4 (−4,−2) −2 (−2,−1) −1 (−1, 2) 2 (2,∞)

10(x − 2) − − − − − − − 0 +x + 2 − − − 0 + + + + +

(x + 1)2 + + + + + 0 + + +(x + 4)2 + 0 + + + + + + +

f ′(x) + ∗ + 0 − ∗ − 0 +

So

f is increasing on the intervals (−∞,−4), (−4,−2) and (2,∞);f is decreasing on the intervals (−2,−1) and (−1, 2);f has stationary points at −2 and 2.

By the First Derivative Test, we deduce that

there is a local maximum at x = −2 with f(−2) = −9;there is a local minimum at x = 2 with f(2) = −1

9 .

6. The denominator is 0 when x = −4 and x = −1, so

the line x = −4 is a vertical asymptote;the line x = −1 is a vertical asymptote.

Thus, by the results of step 4,

f(x) → ∞ as x → −4− and f(x) → −∞ as x → −4+;f(x) → −∞ as x → −1− and f(x) → ∞ as x → −1+.

To find the behaviour of f(x) for large values of x, we divide both thenumerator and denominator of f(x) by the dominant term of thedenominator, x2:

f(x) =1 − 5/x + 4/x2

1 + 5/x + 4/x2, for x �= 0.

Now 1/x → 0 as x → ±∞ and 1/x2 → 0 as x → ±∞, so

f(x) → 1 − (5 × 0) + (4 × 0)1 + (5 × 0) + (4 × 0)

=1 − 0 + 01 + 0 + 0

= 1 as x → ±∞.

Thus

the line y = 1 is a horizontal asymptote,

and

f(x) → 1 as x → ±∞.

30



Exercise 2.7 Sketch the graph of the rational function

f(x) =1

x(x + 1)2.

Finally we use Strategy 2.1 to sketch the graph of a function which is notrational.


f(x) =x√

x2 + 1.

Solution

1. The domain of f is R, since x2 + 1 > 0 for all x in R.

2. f is odd, since

f(−x) =−x√

(−x)2 + 1=

−x√x2 + 1

= −f(x) for all x in R.

3. The solution of f(x) = 0, that is x/√

x2 + 1 = 0, is x = 0, so thex-intercept and the y-intercept are both 0. That is, the graph crossesthe axes only at the origin.

4. Since f(x) has the same sign as x,

f is positive on the interval (0,∞);

f is negative on the interval (−∞, 0).

5. Using the quotient rule, we obtain

f ′(x) =

√x2 + 1 − x(1

2(x2 + 1)−1/22x)x2 + 1

=x2 + 1 − x2

(x2 + 1)3/2=

1(x2 + 1)3/2

. We have multiplied thenumerator and denominator by(x2 + 1)1/2.So f ′(x) > 0 for all x in R; that is,

f is increasing on R.

In addition, f ′(0) = 1, so the graph has slope 1 at the origin. Although finding f ′(0) is notpart of Strategy 2.1, here it iseasy to do and helps with thesketch.

31


6. To find the behaviour of f(x) for large positive values of x, we divideboth the numerator and denominator of f(x) by the dominant term ofthe denominator, x:

f(x) =1√

1 + 1/x2, for x > 0.

So Dividing the denominator by ximplies that, within thesquare-root sign, we divideby x2, and this allows us todetermine the limit as x → ∞.Don’t worry if you find thistricky to follow: you will not beassessed on such a difficultgraph.

f(x) → 1 as x → ∞.

Thus

the line y = 1 is a horizontal asymptote.

Since f is odd,

f(x) → −1 as x → −∞.

Thus

the line y = −1 is a horizontal asymptote.


Earlier in this section we introduced the First Derivative Test to determinewhether a given stationary point gives a local maximum, local minimum orneither. There is an alternative test for a local maximum or local You may have met this test if

you have studied calculuspreviously.

minimum, using the second derivative of the function f .

Second Derivative Test Suppose that a is a stationary point of adifferentiable function f ; that is, f ′(a) = 0. To use this test, f has to be a

twice-differentiable function.1. If f ′′(a) < 0, then f has a local maximum at a.2. If f ′′(a) > 0, then f has a local minimum at a.

This test can be very efficient as a means of classifying stationary points.However, for some functions it is too complicated to find the secondderivative. Moreover, if f ′′(a) = 0, then the Second Derivative Test givesno result; the stationary point may be a local maximum, a local minimum,or neither in this case. This is why the Strategy 2.1 uses the FirstDerivative Test.

Further exercisesSketch the graph of each of the following functions.

Exercise 2.8 f(x) = 15x5 − x3

Exercise 2.9 f(x) =4x + 3x − 7

32

Section 3 New graphs from old

Exercise 2.10 f(x) =2x

x2 + x − 2

Exercise 2.11 f(x) =1√

1 + x2

3 New graphs from old

After working through this section, you should be able to:

(a) sketch the graph of a combination of two functions, one of which isa trigonometric function;

(b) sketch the graph of a hybrid function, whose rule is defined bydifferent formulas on different parts of its domain.

3.1 Further graph-sketching techniquesWe first illustrate some techniques for sketching the graph of acombination of two functions, one of which is a trigonometric function. Asfar as possible, we follow the steps of Strategy 2.1 but in some examples,part or all of some steps are not necessary. In particular, we try to avoiddifferentiating anything difficult! We can also exploit the known features ofthe trigonometric functions, such as the fact that the values of sinx andcos x oscillate (with period 2π) between the values 1 and −1. Because ofthis oscillation it is often convenient to use other simple graphs asconstruction lines. So, for this subsection, we add another step toStrategy 2.1 as follows.

Strategy 3.1 Extended graph-sketching strategy

To sketch the graph of a given function f , determine the followingfeatures of f (where possible), and show these features in your sketch.

1. The domain of f .

2. Whether f is even, odd or periodic (or none of these).

3. The x-intercepts and y-intercept of f .

4. The intervals on which f is positive or negative.

5. The intervals on which f is increasing or decreasing, the nature ofany stationary points and the value of f at each of these points.

6. The asymptotic behaviour of f .

7. Any appropriate construction lines, and the points where f meetsthese lines.

The following example illustrates Strategy 3.1.


f(x) = x sin x.

33

2 Graph Sketching

Documents

Transcript of 2 Graph Sketching