Análisis matemático práctica

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-La función es continua en el intervalo cerrado [1,3] por ser polinómica.

-Es derivable e el intervalo abierto (1,3) por ser polinómica.

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Entonces existe un punto c en el intervalo abierto con derivada nula en dicho punto f ´ (c)=0

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BLOQUE II

1.f(x) = 2x² - 7x + 10.                 [2;5].               a = 2.     b = 5

f’(c) = (f(b) – f(a))/(b -a)

f’(c) =  (4 – 25)/(5-2) = -21/3 = 7

f’(x) = 4x – 7 » f’(c) = 4c -7 = 7      c = 7/2

2. f(x) = x³ + x -1                      [0;2].               a = 0.    b = 2

f’(c) = (f(b) – f(a))/(b -a)

f’(c) = 9 – (-1)/2 = 5

f’(x) = 3x² + 1  » f’ (c) = 3c² + 1 = 5    » 3c² = 4.   c= ±√4/3

3. f(x) = 5 – 4/x                        [1;4].                a = 1.   b = 4

F’(c) = (f(b) – f(a))/(b -a)

f’(c) =  (4 – 1)/(4 - 1) = 3/3 = 1

f’(x) = 4/x² » f’(c) = 4/c²= 1    c = ±2

4. f(x) = 2x – 3      si x < 4.                    [2;6].         a = 2      b = 6

            -x²+10x-19     si  x ≥ 4

F’(c) = (f(b) – f(a))/(b -a)

F’(c) = 5 – 1/(6-2) = 4/4 = 1

Derivamos ambas funciones

F(x) = 2x-3 » f’(x) = 2         no cumple

F(x) =  -x²+10x-19 = -2x + 10   cumple

F’(c) = -2c + 10 = 1 » f’(c)= -2c = -9 » c= 9/2

5. f(x) = 4x² + 4x         si x ≤ 1.               [0;4].        a = 0   b = 4

             -4x³ + 12x².     Si 1 < x < 2

              x³ - 3x² + 20.      Si x ≥ 2

F’(c) = (f(b) – f(a))/(b -a)

F’(c) = 36 – 0/4 = 9

Derivamos las 3 funciones

¹ F(x) = 4x² + 4x » f’(x) = 8x + 4

F’(c) = 8c + 4 = 9 » c = 5/8

² F(x) = -4x³ + 12x² » f’(x) = -12x² + 24x

F’(c) = -12c² + 24c = 9  » f’(c) = 4c² - 8c – 3 » (2c-3).(2c-1) = 0    c1 = 3/2.  c2= 1/2 

³ F(x) = x³ - 3x² + 20 » f’(x) = 3x² - 6x

F’(c) = 3c² - 6c = 9 » f’(x) = c² - 2c – 3 = 0 » (c – 3).(c + 1) = 0      c1 = 3     c2 = -1

6. f(x) = x³ - x²                      [-1;3].                a = -1.         b = 4

f’(c) = (f(b) – f(a))/(b -a)

f’(c) = [18– (-2)]/[(3 – (-1)] = 20/4 = 5

f’(x) = 3x² - 2x » f’(c) = 3c² - 2c = 5 » f’(c) = 3c² -2c -5 = 0

[-b ± √b² - 4ac]/2(a).    =  [ –(-2) ± √(-2)² - 4(3)(-5)] / 2(3)

(2 ± √4+60)/2(3) = (2 ± 8)/6.       C1= 10/6 = 5/3.     C2 = -6/6 = 1

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Entonces aplicando la Regla de L´Hopital se tiene:

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Aplicando L’Hopital:

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BLOQUE V: Hallar los extremos absolutos de las funciones[pic 124]

1.           [pic 125][pic 126]

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 , PUNTOS CRÍTICOS[pic 128]

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