Relaciones entre funciones

Presentación del tema: "Relaciones entre funciones"— Transcripción de la presentación:

1 Relaciones entre funciones
(M) The Mean Value Theorem is a corner stone of calculus. The study of functions depends on the Mean Value Theorem, and many of the fundamental results of calculus need the Mean Value Theorem. (F) In this module we consider one particular application of the Mean Value Theorem. This has to do with proving complicated formulae. We use the Mean Value Theorem to show that certain equalities between functions are true. Aplicaciones de la derivada /Relaciones entre funciones

2 Estimación de Funciones
Las aplicaciones de la derivación para probar igualdades y desigualdades están basadas en una consecuencia del Teorema del Valor Medio: Teorema A Si f es derivable en un intervalo abierto y f’(x) = 0 para todo x, entonces f es una función constante. Teorema B Si f es derivable en un intervalo abierto y f’(x) > 0 para todo x excepto por un numero finito de puntos, entonces f es estrictamente creciente. NOTA: Se supone que f(x) es derivable en un intervalo abierto. Por lo tanto estas conclusiones no se pueden aplicar en casos generales donde la función no esté definida en un intervalo determinado. Por ejemplo, la función f(x) = |x|/x definida para x ≠ 0, es derivable en su dominio, su derivada es nula, pero la función no es constante. (M) The Mean Value Theorem has two important consequences. One is the fact that, if the derivative of a function vanishes, then the function is a constant function. The other is that if the derivative of a function is positive, then the function is increasing. (++) (F) One point has to be observed. What you just said is true only if the function is defined on an interval. For example, the derivative of the function f(x) = absolute value of x divided by x is everywhere 0, but the function is not a constant function. Clearly f(x)=1 if x is positive, and f(x) = -1 if x is negative. The point here is that the function f is not defined for x=0, that is, the domain of definition of the function is the union of two intervals, not a single interval. This point has always to be taken into account when using the Mean Value Theorem. Aplicaciones de la derivada /Relaciones entre funciones

3 Igualdades funcionales(1)
Ejemplos 1 Demostrar que cos2( x ) + sen2( x ) = 1. No hace falta ningún cálculo para demostrar esta fórmula, aunque para obtener muchas fórmulas de derivación se necesita esta igualdad trigonométrica. Pero es recomendable ver como se hace utilizando el cálculo diferencial. Demostración Consideremos la función f( x ) = cos2 x + sin2 x. Derivando obtenemos: f’(x) = 2 cos(x)(-sin(x))+2sin(x)cos(x) = 0. (M) As an example of the uses of the Mean Value Theorem, we show, by Theorem A, that Cos squared plus sin squared always equals 1. (++) calculus is not really needed to prove this formula which follows immediately from the definition of the trigonometric functions. It is, however, instructive to see how calculus methods can be applied here. (++) (F) Consider the function f(x) equals cos squared x plus sin squared x. (F) Straightforward differentiation yields that the derivative of f is two times cos times negative of sin plus 2 times sin times cos. Here the terms cancel out, and the derivative of f is always zero. Hence f is a constant function. (M) To find out which constant f is, simply evaluate it at the point x = 0. Cos at 0 is 1, sin at 0 is 0. We conclude that the value of f at that point is 1. Hence cos squared plus sin squared is always 1. (F) Here it is important that the function f is defined everywhere. That is necessary in order to be able to make the conclusion using the Mean Value Theorem. Por lo tanto f es una función constante. Como f(0) = 1, f(x) = 1 para todos los posibles valores de x. Aplicaciones de la derivada /Relaciones entre funciones

4 Igualdades Funcionales(2)
Ejemplos Demostrar que para x ≥ 0. 2 Demostración Consideremos la función Derivando obtenemos que f’(x) = 0 para todo x ≥ 0. La derivada es algo complicada (podéis usar WIRIS o cualquier otro programa matemático para ello). (M) A more complicated problem is to show that arcsin of x-1 divided by x+1 plus Pi over 2 equals 2 times arctan of square root of x for positive values of x. (F) The assumption that x is positive is necessary, because square roots of negative numbers are not defined. (++) (M) To prove the formula, consider the function f(x) equals arcsin of x-1 divided by x+1 plus Pi over 2 minus 2 times arctan of the square root of x. (++) Straightforward differentiation yields that the derivative of the function f is 0. (F) This is a very unpleasant computation if done manually. You should use a computer algebra system like WIRIS or Maple for this computation. Such systems do all the hard work for you. (M) We conclude then that f is a constant function. Since the value of f at x equals 0 is 0, the formula holds. (F) Here again it is important that the function f is defined on an interval. In this case the function f is defined on the interval from 0 to plus infinity. Hence we were able to make the conclusion. Por lo que podemos afirmar que f es una función constante. Como f(0) = 0 tenemos que f(x) = 0 para todo x. Aplicaciones de la derivada /Relaciones entre funciones

5 Igualdades Funcionales(3)
Ejemplos Demostrar que: Solución Consideremos la función: f(x) = arctg(cotgh(x)) – arccotg(e2x). La función f está definida para x ≠ 0. En x = 0 la función f no está definida porque cotgh(0) no existe. (M) The final example is mean. We have to show that arctan of the hyperbolic cotangent of x minus arc cotangent of e to the power 2 times x equals Pi/4 if x is positive and -3 times Pi over 4 if x is negative. To show the formula, consider the left hand side of the formula. Call it f of x. We have to show that this function f takes one constant value if x is positive and another constant value if x is negative. (++) (F) This function is not defined for x equals 0, because the hyperbolic cotangent is not defined then. Hence we must study this function separately for x positive and for negative. (M) Straightforward differentiation using a computer algebra system shows also here that the derivative of the function f vanishes. (F) Really, use a machine here. This computation is not to be done manually. But it can easily be done with a computer algebra system. Derivando (utilizando el ordenador) obtenemos que f’(x) = 0 para todo x ≠ 0. Aplicaciones de la derivada /Relaciones entre funciones

6 Igualdades Funcionales(4)
Ejemplos Estas constantes C1 y C2 pueden no ser iguales si f no está definida en x=0. Solución Si f’(x) = 0 para la función f(x) = arctg (coth(x)) – arccotg(e2x), el Teorema del Valor Medio implica que: f(x) = C1 una constante para todo x > f(x) = C2 otra constante para x < 0. Para determinar dichas constantes C1 y C2 hay que observar que y que Posteriormente hallamos los limites, teniendo en cuenta que: (M) We now know that the derivative of the function f, that is the derivative of the left hand side of the equation of the problem, vanishes. (++) (M)Hence the Mean Value Theorem tells us that, the function f takes a constant value C one for positive values of x and some other constant value C two for negative values of x. (M) These constants may be different since the function f is undefined for x equals 0. (F) To determine the constants C one and C two observe that the limit of f as x grow towards the positive infinity must be the constant C one. Likewise the limit of f as x goes to the negative infinity must be the constant C two. (M) These limits can be computed directly by the definitions of the functions concerned. The limit, as x goes to the positive infinity of arc tangent of the hyperbolic cotangent of x is Pi over 4, and the limit of the same function as x goes to the negative infinity is negative Pi over 4. Also the limit, as x goes to the positive infinity, of arc cotangent of e to two times x is 0, and the limit of the same function as x goes to the negative infinity is Pi over 2. All this follows from the definition of the functions involved. The ocmbination of these results proves the statement of the problem. (F) This is a rather interesting formula. To prove such a mean formula without using the Mean Value Theorem would be a nightmare. Calculus provides us with some very powerful tools indeed. Aplicaciones de la derivada /Relaciones entre funciones

7 Cálculo en una variable
Traducción al español: Félix Alonso Gerardo Rodríguez Agustín de la Villa Autor: Mika Seppälä