Math symbols
About us
Contact us
Site map
Search The Site
   Program of Lessons
 Study Guide
 Topics of problems
 Tests & exams
www.bymath.com Study Guide - Arithmetic Study Guide - Algebra Study Guide - Geometry Study Guide - Trigonometry Study Guide - Functions & Graphs Study Guide - Principles of Analysis Study Guide - Sets Study Guide - Probability Study Guide - Analytic Geometry Select topic of problems Select test & exam Rules Price-list Registration

Application of derivative in investigation of functions

Continuity and differentiability of function. Sufficient
conditions of functions monotony. Darboux's theorem.
Intervals of function monotony. Critical points.
Extreme ( minimum, maximum ). Points of extreme.
Necessary condition of extreme. Sufficient conditions
of extreme. Plan of function investigation.

Relation between continuity and differentiability of function. If a function  is differentiable at some point, then it is a continuous function at this point. Contrary is invalid: a continuous function can have no derivative.
C o n s e q u e n c e . If a function is discontinuous at some point,then it has no derivative in this point.

E x a m p l e . The function  y = | x |  ( Fig.3 )  is continuous everywhere, but it has no derivative at  x = 0 ,
because a tangent of the graph at this point does not exist.  ( think, please, why ? )

Sufficient conditions of functions monotony.
If   f ’( x ) > 0  at every point of an interval  ( a, b ) , then a function   f ( x ) increases within this interval.
If   f ’( x ) < 0  at every point of an interval  ( a, b ) , then a function   f ( x ) decreases within this interval.

Darboux’s theorem. Points, at which a derivative of a function is equal to 0  or doesn’t exist, divide a function domain for such intervals that within each of them a derivative saves a constant sign.

Using these signs it is possible to find intervals of monotony of functions, what is very important in investigations of functions.



                      Hence, the function increases in the intervals ( - , 0 ) and ( 1, + ) and
                      decreases in the interval ( 0,  1 ). The point  x = 0  isn’t included in the function
                     domain, but as x approaches 0 an item  x - 2   increases unboundedly, therefore
                     the function also increases unboundedly. At the point  x = 1  the function value
                     is 3. According to this analysis  we can draw the graph of the function,
                      represented on  Fig.4b .

Critical points.  A domain interior points, in which a derivative of a function is equal to zero or doesn’t exist, are called critical points of this function. These points are very important at analysis and drawing a function graph, because only they can be points, in which a function has an extreme ( minimum or maximum,  Fig.5a, b ).

At points x1 , x2 ( Fig.5a ) and x3 ( Fig.5b ) a derivative is equal to 0; at points x1 , x2 ( Fig.5b ) a derivative doesn’t exist. But all they are points of extreme.

Necessary condition of extreme. If   x0   is an extreme point of a function  f ( x )   and a derivative  f’  exists at this point, then  f’ ( x0 ) = 0.

This theorem is a necessary condition of extreme. If a derivative of a function at some point is equal to zero, then it’s not necessarily, that the function has an extreme at this point. For instance, a derivative of the function  f ( x ) = x³  is equal to 0  at x = 0 , but this function has no extreme at this point  ( Fig.6 ).

On the other hand, the function y = | x | , represented on Fig.3, has a minimum at the point x = 0 , but there is no derivative at this point.

Sufficient conditions of extreme.  If a derivative changes its sign from plus to minus at a point  x0 , then  x0  is a point of maximum.  If a derivative changes its sign from minus to plus at a point  x0 , then  x0  is a point of minimum.

Plan of function investigation. To draw a graph of a function it is necessary:

    1)  to find a domain and a codomain of a function,

    2)  to ascertain if the function is even or odd,

    3)  to determine if the function is periodic or not,

    4)  to find zeros of the function and its values at  x = 0,

    5)  to find intervals of a sign constancy,

    6)  to find intervals of monotony,

    7)  to find points of extreme and values of a function in these points,

    8)  to analyze the behavior of a function near “special” points and at a great modulus of x .

E x a m p l e . Analyze the function  f ( x ) = x³ + 2x² - x - 2 and draw its graph.

S o l u t i o n . We’ll investigate the function by the above represented scheme.

                      1)  a domain x R ( i.e. x – any real number ); a codomain  y R, because
                            f ( x )  is an odd degree polynomial;
                      2)  f ( x ) is neither an even nor an odd function ( explain, please );
                      3)  f ( x ) is not a periodic function ( prove this, please, yourself );
                      4)  a graph of this function is intersected with  y-axis in the point ( 0, - 2 ),
                             because  f ( 0 ) = - 2 ;  to find zeros of the function it is necessary to solve
                             the equation:  x³ + 2x² - x - 2  = 0 , one of roots of which ( x = 1 ) is
                             obvious. Other roots can be found ( if they exist ! ) by solving the quadratic
                            equation: x² + 3x + 2 = 0, which is received after dividing the polynomial
                             x³ + 2x² - x - 2  by the binomial  ( x – 1 ). It is easy to check, that the
                            two other roots are: x2 = -2 and  x3  = -1. So, zeros of this function are:
                              -2, -1 and 1.
                       5)  It means, that a numerical line is divided by these roots into four intervals of
                            a sign constancy, inside of which the function saves
                             its sign :

                            This result can be received from the polynomial factorization:

                                           x³ + 2x² - x - 2 = ( x + 2 ) ( x + 1 ( x – 1 )

                             and estimating the product sign by the method of intervals.
                       6)  The derivative:  f’ ( x ) = 3x² + 4x -1  has no points, at which it doesn’t
                            exist, therefore its domain is R ( all real numbers ); zeros of f’ ( x ) are
                             roots of the equation:  3x² + 4x - 1 = 0 .


                             The received results are included in the following table:


| Home | About us | Links | Contact us |

Copyright © 2002-2012 Dr. Yury Berengard. All rights reserved.