Application of derivative in investigation of functionsContinuity 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: