Maths6

Országok listájaHungaryBudapesti Corvinus EgyetemGazdálkodástudományi KarNemzetközi gazdálkodás (angol nyelven)Mathematics1JegyzetekMaths6

Graphing of functions
6.1
that

Finding extreme values
f
is dierentiable at

f : R R and let x0 be an interior point of its domain. Suppose x0 . We saw that the condition f (x0 ) = 0 is necessary for the function to have an extreme value at x0 . We are going to nd sucient conditions for the existence of extreme values. It is clear that if we can nd a number > 0 such that f is decreasing on the interval [x0 - , x0 ] and increasing on [x0 , x0 + ] then f has a local minimum at x0 .
Consider a function For dierentiable function we formulate this observation in the following theorem.

Theorem 6.1 Suppose that f is dierentiable on an interval and x0 is an interior point of this interval. If there exists a number > 0 such that
· f (x) 0 , · f (x) 0 ,

if x (x0 - , x0 ) if x (x0 , x0 + )

then f has a local minimum at x0 .
Obviously, an analogous statement holds for maximum values.

Example 6.1

Find the extreme values and intervals of monotonicity of the function

f (x) = x2 e-x .
It can be seen that the sign of the derivative

f (x) = (2x - x2 )e-x
depends only on the rst term. Hence we get.

· · · · · x=0

If if if if if

x (-, 0), x = 0,
then

then

f (x) < 0,

therefore

f

is decreasing;

f (0) = 0,

this is a critical point; therefore

x (0, 2), x = 2,

then

f (x) > 0,

f

is increasing;

then

f (2) = 0,
then

this is a critical point; therefore

x (2, +),

f (x) < 0,

f

is decreasing.

Considering the changes of sign of and has a local maximum at

f we see x = 2.

that

f

has an (absolute) minimum at

1

6.2.

SECOND-ORDER CONDITIONS

Example 6.2

Consider the function

f (x) = x + sin x
on the real line. Since points at

f (x) = 1 + cos x, x = (2k + 1)

it is clear that the function has critical

k = 0, ±1, ±2, . . . .

At non of these points can the function have an extreme value, since for these points we have

cos x > -1,

consequently

x = (2k + 1)

for

f (x) > 0 .

Thus, this function is strictly increasing on the whole real line.

6.2

Second-order conditions
In such cases the following second-order

It can occur that we have to investigate complicated functions, where it is dicult to determine the sign of the derivative. (sucient) condition for the extreme value can be useful.

Theorem 6.2 Let f be dierentiable on some interval and suppose that in some interior point x0 of the interval f is twice dierentiable. If f (x0 ) = 0 and f (x0 ) > 0 then f has a local minimum at x0 .
Indeed, investigating the dierence quotient we see that

f (x0 )

f (x0 + h) - f (x0 ) = h f (x0 + h) >0 = lim h0 h =
h0

lim

which means that the quotient

f (x0 + h)/h

is positive, if

0 < |h| <

for some

> 0. · ·

Hence

if if

x (x0 - , x0 ), x (x0 , x0 + ),

then then

f (x) < 0, f (x) > 0. f
has a local minimum at

Therefore, by the 6.1 Theorem

x0 ,

as stated.

Of course, an analogous sucient condition an be formulated for local maxima. By an indirect argument we get the second-order necessary condition for the extreme value.

Theorem 6.3 Suppose that f is twice dierentiable on an interval and let x0 be an interior point of the interval.
· ·

If f has a local minimum at x0 then f (x0 ) = 0 and f (x0 ) 0. If f has a local maximum at x0 then f (x0 ) = 0 and f (x0 ) 0.
Let

Example 6.3
for

f (x) = x ln x x > 0. Then f (x) = 1 + ln x, thus x = 1/e. Since f (x) = 1/x, we have
the only critical point of the function

f

is

f (1/e) = e > 0 .
Therefore

f

has a local minimum at

x = 1/e.

It is not dicult to see that this is

the absolute minimum.

2

6.3.

CONVEX AND CONCAVE FUNCTIONS

Observe that the theorems we proved so far do not say anything about a critical point

x0

for which

f (x0 ) = 0 .
The reason is that in this limiting case anything can occur. For instance, consider the behavior of the power function

f (x) = xn
in the critical point hand,

(n 3) f (0) = 0
and

x0 = 0. f

It is clear that

f (0) = 0.

On the other

· ·

if if

n n

is even then is odd then

has an (absolute) minimum at

x0 = 0,

f (x0 ) = f (0) -f

is not an extreme value.

Analogously, if

n

is odd then

has an (absolute) maximum at

x0 = 0.

6.3
x1
and

Convex and concave functions
The function

Denition 6.1
x2

f

is convex on the interval

of the interval and for any real number

[a, b] 01

if for any two points

f (x1 + (1 - )x2 ) f (x1 ) + (1 - )f (x2 )
holds. Geometrically this means that the points of thesecant line are not below the points of the graph of the function in the

(x1 , x2 )

interval.

A concave function is dened by the reversed inequality. We give a simple and geometrically obvious characterization of convexity for twice dierentiable functions.

Theorem 6.4 Assume that the function f is continuous on some interval and is twice dierentiable in the interior of the interval. The sucient and necessary condition for f being convex on the interval is that
f (x) 0

for each interior point of the interval.
This means that for a convex function the slope of the tangent lines to the graph of the function increases. An other way of visualizing this is that the graph of the function is nowhere below the tangent line to the graph.

Example 6.4 Consider the function
f (x) = x . 1 + x2

It is easy to check that
f (x) =

1 - x2 . (1 + x2 )2

Investigating the sign of the derivative it can be seen that
· f · f · f

strictly decreasing on the interval (-, -1), has an (absolute) minimum at x = -1, strictly increasing on the interval (-1, 1),
3

6.3.

CONVEX AND CONCAVE FUNCTIONS

· f · f

has an (absolute) maximum at x = 1, strictly decreasing on the interval (1, +).
2x3 - 6x . (1 + x2 )3

We check the convexity based on the sign of the second derivative
f (x) =

It is clear that the denominator is everywhere positive, therefore it is enough to consider the sign of
2x3 - 6x = 2x(x2 - 3) .

Thus
· f · f · f · f

is concave on the interval (-, - 3), is convex on the interval (- 3, 0), is concave on the interval (0, 3), is convex on the interval a ( 3, +) .


Observe that f (- 3) = f (0) = f ( 3) = 0, and the second derivative changes its sign in these points. These points separate the intervals where the function is convex or concave, respectively. Such points are called inection points of f.
In inection points the tangent line intersects the graph of the function. One of the most important properties of convex functions is that a local minimum is at the same time the absolute minimum of the function.

Theorem 6.5 Consider a function f that is dierentiable twice and convex on an interval and let x0 be an interior point of the interval. If f has a local minimum at x0 , then this is the absolute minimum of the function on the interval.
Indeed, we have due to the convexity of

f (x0 ) = 0 by assumption, on the other hand, f f . Hence we get for the interior points of the

is increasing interval that

· f (x) 0 · f (x) 0

if if

x < x0 , x > x0 . f.

The statement now follows by the monotonicity of concave functions.

Of course, an analogous statement can be formulated for maximum values of

Procedure for sketching the graph of a function f .
x-intercepts
and

1. Determine the domain. 2. If possible nd the

y -intercept.

3. Determine if the function is periodic, even or odd. 4. Find the derivative and the critical values of zero, and the points where

f.

(where the derivative equals

f

is not dierentiable but continuous).

5. Determine the intervals where trema.

f

is increasing/decreasing and classify the ex-

6. Find the second derivative and the points where it is zero.

4

6.3.

CONVEX AND CONCAVE FUNCTIONS

7. Determinethe the concavity and locate the inection points. 8. Determine the limits at the edges of the domain. 9. Sketch the graph. 10. Determine the range.

Exercise 6.1 Carry out the whole curve sketching procedure for the following functions: 1) y = x3 - 3x2 + 3x + 1 2) y = 2x2 - x3 3 4 3) y = x + x3 - 3x2 + 3 4) y = 2x2 - x4

5) 7) 9) 11) 13) 15) 17)

x2 (x - 1)2 1 y =x+ x-3 4 y = -x x y= y = x - arctan x y = x ln x y = (1 - x)3 (1 + x)3 1 y = xe- x

4

6) 8)

y = x + sin x y =x+ 1 x2 2 4 - x2 = x 4x = 16 + x2 = ex - e-x = x3 (x2 - 1) = x-x

10) y 12) 14) 16) 18)
y y y y

5



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