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7/28/2019 CD_U4_A3_ERMR
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ESCUELA SUPERIOR ABIERTA Y A
DISTANCIA
CALCULO DIFERENCIAL
UNIDAD 4 ACTIVIDAD 3:Aplicacin del criterio de la primera derivada
ERIK MTZ
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AL12529616
a) en el intervalof(x) = x2 x = 2x -1 =02x =1 o x = 2x =1 si x =-y cuando x = f(x) = 2x 1f(-) = 2(-) 1= -1f() = 2() 1= 1(- , ) decreciente( , ) crecienteLos puntos crticosx=0 ,x = ,x=- y x =
b) en el intervalo
f(x) = x3 x2 = 3x2 2x =0 Los puntos crticos3x2 = 2x x = x=0
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f(x) = 3x2 2x x= f(-) = 3(-)2 2(-)= f() = 3()2 2()= - x = - y x = (-, ) decreciente( , ) creciente
c) en el intervalo
f(x) = x2ex = ex(x2 +2x) =0x2 = 0 o x =
2x = 0 o x = f(x) = ex(x2+2x)f(-3) = ex((-3)2+2(-3)) = f(-3) = ex(9 - 6) = 3 exf(3) = ex((3)2+2(3)) =f(3) = ex(9 + 6) = 15 ex
(-3, ) decreciente( , 3) crecienteLos puntos crticosx = 0,x = x =
d) en el intervalof(x) = sen2(x) = 2cos(x) sen(x) =0cos(x)=0 o sen(x)=0x = 0 o x = cos(x) = 0 x = x = f(x) = 2cos(x) sen(x)f(0) = 2cos(0) sen(0) =f(0) = 0 f(2) = 2cos(2) sen(2) =f(2) = cos(4) sen(4)(0, 0) decreciente
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(cos(4) sen(4) , 2) crecienteLos puntos crticosx=0,x=cos(x)=0x = x =
e) f(x) = e(3x+5)2 = e(3x+5)22(3x+5)(3)f(x) = e(3x+5)2 6(3x+5)= e(3x+5)
2(18x + 30) = 0x = 0 x = -
f(x) = e(3x+5)2 (18x + 30)f(-2) = e(3x+5)2 (18(-2) + 30) = -6f(2) = e(3x+5)2 (18(2) + 30) = 66
(-2, -6) decreciente(2 , 66) crecienteLos puntos crticosx =0,x = ,x = -2x = 2
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