Resolución Detallada de Problemas de Cálculo Diferencial e Integral
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f(x)=x²-8lnx f'(x)derivabl n(0,∞) f'(x)=2x-(8/x)=0→x=-2(no) x=2
f'(1,5)=2(1,5)-(8/1,5)=-2,3<0 decrec="">0>
f'(2,5)=2(2,5)-(8/2,5)=1,8>0 crecient (2,e)
f(1)=1-0=1 f(e)=0,61 f''(x)=2+(8/x²) f''(2)=2+2=4>0convex(1,e)
f(x)=x³-4x f(1)=1³-4=-3 f'(x)=3x²-4 f'(1)=3(1)²-4=-1
y+3=-1(x-1)=x+1 y=-x-2 limf(x)x→-∞=-∞ limf(x)x→+∞=+∞
f(x)=0 x=0(0,0) x=+-2(-2,0)(2,0)
x³-4x=-x-2
x³-3x+2=0 x=1 y -2
y=x-2 x=1 y 2
∫¹-2[(x³-4x)-(x-2)]dx=∫¹-2(x³-3x+2)dx=[x4/4 - 3x²/2 +2x]¹-2=27/4u²
f(x)=ex(x²-x+1) Limx→-∞ ex(x²-x+1)=Limx→+∞ x²+x+1/ex=∞/∞
L'H limx→+∞ 2x+1/ex=∞/∞ L'H Limx→+∞ 2/ex=0 Limx→+∞f(x)=∞
f'(x)=0 x²+x=0=x(x+1) x=0 x=-1 posibles extrem relat
f'(-2)=>0 crecient (-∞,-1) f'(-0,1)=<0 decrecient="">0>
f'(1)>0crecient(1,+∞) x=-1 max relat(3/e) x=0 mín relat(1)
f'(x)=ex(x²+x) f''(x)=ex(x²+x)+ex(2x+1)=ex(x²+3x+1)
f''(-3)=>0 ∩(-∞,-2,6)f''(-1)=<0 u(-2'6,0'38)="">0>''(0)=>0∩(-0'38,∞)
f(x)=x²-2x g(x)=-x²+4x f(x)=g(x) 2x²-6x=0=x(2x-6) x= 0 y 3
∫³0[(-x²+4x)-(x²-2x)]dx=
=∫³0(-2x²+6x)dx=
=[-2x³/3 + 3x²]³0=
=-18+27=9u²
f V(1,-1) g V(2,4)
f(x) { x+k si x≤0| ex²-1/(x²) si x>0 f(0)=limf(x)x→0-=limf(x)x→0+
f(0)=(0+k)=k limx→0+ ex²-1/x²=0/0 L'H limx→0+ 2x·ex²/2x=e0=1
k=1 y - f(1)=f'(1)(x-1) f(1)=e¹-1 f'(1)=2 y - (e¹-1)=2(x-1)
I=∫¹0 x/(1+√1-x) dx I=∫(1-t²)/(1+t) 2tdt=
∫[(1-t)(1+t)]/(1+t)](-2t)dt =-2∫(t-t²)dt=-2(t²/2 - t³/3)=-t²+(2/3)t³=
-(√1+x)²+ 2/3(√1-x)³ I=∫¹0 x/(1+√1-x)dx=[-(1-x)+2/3(√1-x)³]¹0=
=(-1(1-1)+ 2/3(√1-1)³)-(-(1-0)+ 2/3(√1-0)³=0+1 - 2/3=1/3
f(x)=e-x/1-x x≠1 limf(x)x→1-=+∞ limf(x)x→1+=-∞ x=1 A.V.
limf(x)x→+∞=0 y=0 A.H(+∞)limf(x)x→-∞=∞/∞ → L'H → =+∞
no A.H(-∞) f'(x)=(x·e-x/(1-x)² f'(x)=0 x=0
f'(-1)=<0>0> f'(2)=>0 creciente (0,+∞)-{1}
f(x)=9-x²/4 f(1)=8/4=2 f'(1)=-1/2 y=-x/2 + 5/2 (1,2) (5,0)
(x=1, y=2) (x=5, y=0) ∫³1 (9-x)²/4 dx=
4-[9x/4 - x³/12]³1=4-[(27/4 - 27/12)-(9/4-1/12)]=5/3 u²