Higher_Mathematics_Part_1

Conclusion. If a function has the relative extremum at the point, then this point is critical. But not every critical point is a point of extremum.

Theorem 3.18	(first sufficient condition of relative extremum). Let x0 be
	critical point of the function f (x) . Suppose there exists neigh-
bourhood (x0 − δ; x0 + δ)		of	point x0 , in which	a function has a	derivative
f ′(x) except, probably, point			x0 . Then:	f ′(x) > 0 and on the interval
1) if on the interval (x0 − δ; x0 ) the derivative
(x0 ; x0 + δ) the derivative f ′(x) < 0 then the point x0 is the point of relative
maximum of the function		f (x) ;
2) if on the interval (x0 − δ; x0 ) the derivative				f ′(x) < 0 and on the interval
(x0 ; x0 + δ) the	derivative f ′(x) > 0 , then the point x0 is the point				of relative
minimum of function f (x) ;
3) if on both intervals		(x0 − δ; x0 ) and (x0 ; x0 + δ) the derivative			f ′(x) has
the same sign, then the point			x0 is not the extreme point of function f (x) .

In other words, if when passing the critical point x0 from right to left the sing of the derivative f ′(x) has changed from plus to minus, then x0 is the

point of relative maximum, if the sing of the derivative changes from minus to plus, then point x0 is point of relative minimum, if the derivative does not

change the sing, then in point x0 extremum is absent.

The rule for investigation of extremum of the function

For finding relative extremum of function f (x) it is needed:

1) to find the critical points of the function f (x) . For this it is necessary to solve the equation f ′(x) = 0 and from its solutions we must to select that,

which belongs to the domain of existence of function; find points in which the derivative does not exist;

2)if the critical point exists, it is needed to investigate the sing of derivative at all intervals, on which the domain of existence is divided by these critical points. For this purpose it is enough to determine the sign at any point of this interval, because the derivative can change its sign only when passing the critical point;

3)after changing the sign f ′(x) when passing the critical points from left

to right to define the points of maximum and minimum. When it is necessary to define the meaning of the function f (x) in these points.

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Theorem 3.19	(the second sufficient condition for relative extremum). Let
	x0 be fixed point of the function f (x) , so f ′(x0 ) = 0 and in

neighbourhood of point x0 there exists second derivative, so that f ′′(x0 ) ≠ 0 . If f ′′(x0 ) > 0 , then x0 is the point of relative minimum, if f ′′(x0 ) < 0 , then x0 is the point of relative maximum.

22.2 The greatest and the least values of the function

The most important is not to confuse the local maximum (minimum) with the greatest (least) value of the function, which reaches on a definite segment. The function may have several local maximum and minimum , when the greatest value (it can also be named as absolute maximum), if it exists, unique. The same is true about the least value (absolute minimum) of the function.

To find the greatest and the least values of the function y = f (x) on the interval [a; b] it is necessary:

1)to find the derivative f ′(x) and determine the critical points of the given function;

2)to calculate the value of the function on those critical points, which belong to the interval (a; b) , and also in points a and b ;

3)to select between the obtained values the greatest and the least values.

22.3. The concavity of the curve. The inflection points

Definition 3.27. Curve y = f (x) is called concave down on the interval (a; b) , if all its points, besides the tangent point, lie below its any tangent lines on the interval (Fig.3.25).

Definition 3.28. The curve y = f (x) is called concave up on the interval (a; b) , if all its points, besides the tangent point, lie above its any tangent lines

on the interval (Fig.3.26).

The inflection point is such a point of the curve, which marks out its parts of concavity (Fig.3.27).

y	М	y		y
			М			М
О	а	b x О	а	b x О	а	b x
	Fig. 3.25		Fig. 3.26			Fig. 3.27

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For investigation of the graph of a function on the intervals for concavity the second derivative of a function is used.

Theorem 3.20		Let function y = f (x) be twice	differentiable on	interval
		(a; b) . Then:
1) if	f ′′(x) < 0 , x (a; b) , then the graph of function y = f (x) is concave
down on interval (a; b) ;
2) if	′′	> 0 , x (a; b) , then the curve y	= f (x) is concave	up on
	f (x)

interval (a; b) .

From the theorem it comes, that at the inflection point the second derivative is equal to zero, if it exists. But the inflection points of the curve y = f (x) may

be the points, in which the second derivative f ′(x) does not exist (for example point x = 0 of the curve f (x) = 5 x ).

Points in which the second derivative f ′(x) is equal to zero or does not exist, are called the critical points of the second type of the function y = f (x) . So, if x0 is abscissa of inflection point, then x0 is critical point of the second

type of this function. The opposite statement is incorrect.

Let’s formulate sufficient conditions of existence of inflection point.

Theorem 3.21 Let x0 be the critical point of the second type of function f (x) . If passing the point x0 f ′′(x) changes its sign, then

point (x0 ; f (x0 )) is the inflection point of the curve f (x) .

22.4. Asymptotes of the curve

There exist three types of asymptotes: vertical, inclined, horizontal. The straight line x = c is vertical asymptote (Fig. 3.28), if

lim			f (x) = ±∞		or	lim	f (x) = ±∞ .
x → c + 0						x → c − 0
The straight line	y = kx +b is inclined asymptote (fig.3.29), if the following
finite limits exist
lim		f (x)		= k (k ≠ 0),		lim ( f (x) − kx) = b .
x → ∞			x			x → ∞
Remember, if it	is		necessary to		consider		the case as x → +∞ so and
x → −∞ .
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When lim	f (x) = b ( lim f (x) = b) , then straight line y = b is called
x → +∞	x → −∞

a horizontal asymptote of graph of a function y = f (x) (Fig.3.30).

It is clear, that horizontal asymptote is separate case of inclined asymptote ( k = 0 ).

y		x	y		y
					b	y = b
				y = kx + b
О	с	х	О	х	О	х
	Fig.3.28		Fig.3.29		Fig.3.30

22.5. Scheme of investigation of the graph. Constructing the graphs of functions

For investigation of the function and constructing the graph it is necessary:

1)to find the domain of existence of a function;

2)to find (if it is possible) the points of intersection of a graph with the coordinate axes;

3)to test a function on periodic, odd and even check. It should be noted, that graph of even function is symmetric relatively to axis of ordinates, and graph of odd function is symmetric relatively to the origin;

4)to find the points of discontinuity and set there characteristics;

5)to find the monotony intervals, the local extremum points and the value of a function at these points;

6)to find the intervals of convexity and concavity, and the inflection points;

7)to find the asymptote of the curve;

8)to investigate the behavior of a function in infinite remote points;

9)to calculate if it is necessary the value of a function in several control

points;

10)to sketch the graph of a function.

Micromodule 22

EXAMPLES OF PROBLEMS SOLUTION

Example 1. Find intervals of increasing and decreasing of the function

	f (x) =	x2	+1	.
		(x −1)2

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Solution. The domain is (−∞;1) (1;+∞) . Find the derivative

	f ′(x) =	2x(x − 1)2 − 2(x − 1)(x 2 + 1)	= −		2(x +1)	.
		(x − 1)4
					(x −1)3
	The derivative f ′(x) is equal to zero at the point			x = −1and it doesn’t exist

if x =1 . So, x =−1; 1 are critical points of a given function.

Let’s denote these points at the numerical axis (and we should remember about the domain of a function) and define the sign of derivative at each interval (Fig. 3.31):

f ′( x) :	–		+		–
					x
f ( x) :	-1		1

Fig. 3.31

In such case, the function decreases, if x (−∞;1) (1;+∞) , and increases at the interval (−1; 1) .

Example 2. Find the local extremums of the function f (x) = x −x5 / 5 .

Solution. The domain is (−∞; ∞) . Let’s find the critical points:

′	− x		4	′	if	(1 + x)(1 − x)(1 + x	2	) = 0 x = ±1.
f (x) = 1				, f (x) = 0
So, points		x = ±1 are the				critical (stationary)		points. Define signs of a

derivative at the intervals (Fig. 3.32):

f ′( x) :	–		+		–
					x
f ( x) :	-1		1

Fig. 3.32

As we can see on Fig. 3.32 within the intervals (−∞;−1) , (1;+∞) the function decreases but at the interval (−1; 1) it increases. By the theorem 3.18 we can make a conclusion that x =−1is a point of relative minimum; x =1 is a point of relative maximum, and ymin = y(−1) = −4 / 5 , ymax = y(1) = 4 / 5 .

	Example 3. Find the relative extremums of the function f (x) =	3	x2	.
		x	+2

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Solution. The domain is (−∞;−2) (−2;+∞) . Let’s find the critical points:

2

−1 / 3

x

(x + 2) − x2 / 3

2x

−1/ 3

(x +

2) − 3x

2 / 3

f ′(x) =

3

=

=

(x + 2)2

3(x + 2)2

=

2x + 4 − 3x

= −

x − 4

.

33 x (x + 2)2

33 x (x + 2)2

Equation f ′(x) = 0 has a unique root x = 4 . The derivative doesn’t exist at

the points x = −2 and		x = 0.	Thus,	at the	point			x = −2 the function is
undefined, but at the point		x = 0 it is defined. Let’s define signs of derivative at
each interval (Fig. 3.33):
f ′( x) :		–	–	+				–
f ( x) :		–2		0	4			х
			Fig. 3.33
With conversion through the point				x = 0	from the left to the right, the
derivative changes the sign from minus to plus. On the interval (−2; 0) the
function decreases, but on the interval				(0; 4) the function increases. Therefore

x = 0 is the point of relative minimum. By analogy we are sure, that x = 4 is the

point of relative maximum. The point	x = −2 isn’t the critical point (at this
point the function isn’t defined).
Example 4. Investigate the function	y = 3x4 −4x3 −12x2 +2 on extremum.
Solution. The domain is (−∞;+∞) . The derivative of the given function		is
y′ = 12x3 − 12x 2 − 24x. Solve the equation f ′(x) = 0 :
12x3 −12x2 −24x = 0 ; 12x(x2 −x −2) = 0 ;
x1 = −1 , x2 = 0 , x3 = 2 are stationary points.
The second derivative is y′′ = 36x 2 − 24x − 24 = 12(3x 2 − 2x − 2).		We
define the sing y′′ at the stationary points:
y′′(−1) = 12(2 + 2 − 2) = 36 > 0 , y′′(0) = −24 < 0, y′′(2) = 72 > 0.

By the theorem 3.19 we can make a conclusion that x1 = −1 and x3 = 2 are points of relative minimum and x2 = 0 is the point of relative maximum.

Example 5. Investigate the function f (x) = sin2 x −x2 +1 on extremum at the point x = 0 .

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. This point doesn’t belong

Solution. We have:

f ′(x) = sin 2x − 2x,	f ′(x) = 0 ;
f ′′(x) = 2 cos 2x − 2,	f ′(x) = 0 ;
f ′′′(x) = −4 sin 2x,	f ′(x) = 0 ;
f (4) (x) = −8cos 2x,	f (4) (0) = −8 < 0 .

So, the given function at the point x = 0 has the relative maximum. Example 6. Find the greatest and the least values of the function y = x2 ln x

on the interval [1; e] .

Solution. Let’s find the derivative

y′ = 2x ln x + x = x(2 ln x + 1).

As far as the function is defined when x > 0 , so the critical point we find

from the condition 2 ln x +1= 0 , so x = e−0.5 = 1 e

to the interval [1; e] . So let’s evaluate only the values of the function at the ends

of interval. We have:

y(1) = 0,

y(e) = e2 .

Therefore,

max

f (x) = f (e) = e2 , min

f (x) = f (1) = 0.

x [1;e]

x [1;e]

Example 7. Find the greatest and the least values of the function y =

x +1

x2 +1

on the interval[0; 3] .

Solution. Let’s find the critical points:

′

x 2 + 3 − 2x(x + 1)

− x 2 − 2x + 3

y

=

(x

2 + 1)2

=

(x 2 + 1)2 .

−x2 −2x +3 = 0 , x = −3 , x =1 .

1

2

The point

x1 = −3

is not within the interval[0; 3] . We count the values

f (x ) = f (1)

=

1

;

f (0) =

1

;

f (3) =

1

. That is,

2

2

3

3

max f (x) = f (1) = 1 ,

min

f (x) = f (0) = f (3) =

1 .

x [0;3]

2

x [0;3]

3

Example 8. Find intervals of convexity and concavity and inflection points of the curve f (x) = 3x4 −4x3 +1 .

Solution. Let’s find derivatives:

f ′(x) = 12x3 − 12x 2 ,

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(−∞;1) (1;+∞)..

		2				2
	f ′′(x) = 36x		− 24x = 36x x −				.
						3
				2					2
	Solve the equation f ′′(x) = 0,		36x(x −		) = 0 . So, x = 0 ;			x =		are
			3				1	2	3

critical points of the second type. We define the sign of the second derivative: if

x < 0 , then f ′(x) > 0 and the curve is concave up;					if x is within the interval
(0,	2/3),			then f ′(x) < 0 and the curve is concave	down; if x > 2 / 3 ,	then
f ′(x) > 0 and the curve is concave up. At transition through points x1 = 0						and
x	=	2		the second derivative changes its sign. It follows that points (0; f (0))
2	3

and (2 / 3; f (2 / 3)) ,( so (0;1) and (2 / 3;11/ 27) ) are the inflection points of the given curve.

Example 9. Find the asymptotes of the curve y = x3 + x2 +1 . x2

Solution. Let’s write the equation of the given curve as following:

y = x +1+ 1 . x2

We find the equation of inclined asymptote in such a form y = kx +b . We receive

	k = lim	f (x)	= lim	x + 1+ 1 x2	= lim (1	+ 1	+	1	) = 1;
		x		x				x3
	x→∞		x→∞		x→∞	x

b = lim ( f (x) − kx) = lim (x + 1+ 1 x2		− x) = lim (x + 1 x2 ) = 1.
x→∞	x→∞	x→∞

So, y = x +1is the equation of inclined asymptote. Further, as the function

	y = x +1+	1	at the point x = 0 has the discontinuity of the second type, then
		x2

	the straight line x = 0 is the vertical asymptote of the given curve. The curve
	doesn’t have the horizontal asymptote (lim(x + 1 + 1 x2 ) = ∞).
	x→∞
	Example 10. Investigate the function y =	x2	+1		and construct its graph.
		x	−1

Solution. 1) The domain of existence is the whole numerical axis, except the point x =1 , so

2) The graph of the function y = f (x) intersects the axis of ordinates (if it’s possible) at the point (0; f (0)) . We find y(0) = −1 , so A(0;−1) is the point of

258

intersection of the curve with the axis Oy . To find points of intersection of the

	graph with the axis Ox , it is necessary to solve the equation y = 0 ,	x2	+1	= 0 .
		x	−1

This equation doesn’t have real roots, so given function doesn’t intersect the

abscissa axis.

3) The function

is none-periodical. Let’s consider the expression

f (−x) =

(−x)2

+1

=

x2 +1

.

−x −1

−x −1

Therefore, f (− x) ≠ f (x)

and f (− x) ≠ − f (x) . It

means

that

the given

function is neither even nor odd.

x =1 has discontinuity of the second type, and

4) The function at the point

lim

x2

+ 1

= −∞,

lim =

x2 + 1

= +∞.

In

all

other

points

the

function is

x

− 1

x −1

x→1−0

x→1+0

continuous.

5) We find the derivative

y′

=

2x(x − 1) − (x

2 + 1)

=

x2 − 2x − 1

and solve

(x

− 1)2

(x − 1)2

the equation

y′ = 0 , or x2 −2x −1 = 0 . We get stationary points

x = 1−

2

1

and

x2 = 1+

2 .

Besides,

the

derivative

is

undefined when

x =1 .

So,

x1 = 1−

2 ,

x2 = 1+

2 ,

x3 =1are critical

points

or

points of probable

extremum. These points dissect a numerical axis on four intervals (−∞; 1−

2) ,

(1− 2; 1) ,

(1; 1+

2) ,

(1+

2; ∞) .

On

each of

these

intervals

the

derivative y′

has particular sign, which we can put by the method of intervals or

calculation of values of the derivative in different points (in one point from each interval). On intervals (−∞; 1− 2) , (1+ 2; ∞) the derivative is positive, so

the function increases; if x (1− 2; 1) (1; 1+ 2), then the function decreases, because on these intervals the derivative is negative. At transition through the point x1 = 1− 2 (from the left to the right) the derivative changes

the sign from plus to minus, so, at this point the function has the relative maximum. Then

	ymax = y(1−	2) =	(1−		2)2 + 1	= 2	− 2 2 .
			1	−	2 −1
	At transition through the point x2 =			1+	2 the derivative changes the sign
	from minus to plus, so, at this point there			is relative minimum, and
							259

ymin = y(1+

2) =

(1+

2)2 + 1

= 2 + 2 2 .

1+

2 −1

The point

x = 1 isn’t the point of extremum (at this point the function isn’t

defined).

6) Find the second derivative

x2

− 2x −1 ′

(2x − 2)(x −1)2

− (x2 − 2x −1) 2(x − 1)

4

y′′ =

=

4

=

.

2

(x −1)

(x − 1)

3

(x − 1)

On the interval

(−∞; 1) , y′′ < 0 , sо, on this interval the curve is concave

down; if x (1; ∞) , then

y′′ > 0 and the curve is concave up. At the point x =1

the function isn’t defined, so this point isn’t the inflection point.

7) From the results of item 4 comes up, that the straight line x =1 is vertical asymptote of the curve.

	As lim	=	x2	+ 1		= +∞, lim	x2	+ 1		= −∞,
			x	− 1			x	− 1
	x→+∞					x→−∞

then horizontal asymptotes are missed. Let’s find the limits

lim

f (x)

=

lim

x2

+ 1

= 1, lim

f (x)

=

lim

x2

+ 1

= 1.

x

x

− x

x→+∞

x→+∞ x2 − x

x→−∞

x→−∞ x2

So,

k = 1,

y

2 + 2 2

1

b = lim ( f (x) − kx) =

x→±∞

В•

•y = x + 1

lim (	x2 + 1		− x) = lim	x + 1		= 1.
x→±∞ x − 1			x→±∞ x −1

In such a way, the straight y = x +1

is the inclined asymptote of the given curve. There are no other asymptotes.

8) Let’s calculate a few values of the function: y(3) = 5 , y(−1) = −1.

So, points В(3;5), С(–1;–1) belongs to the graph.

9) Taking into account the spent examinations, we draw the graph (see Fig. 3.34).

			• А 1 1+		x	Example 11. Investigate the func-
–1		•		2
•						tion y = 3 x3 −2x2 and draw the graph
	С
						of this function.
						Solution. 1) The domain is whole
			Fig. 3.34			axis, so D( y) = (−∞; +∞).

260