a y b   fx=ax²+blnx   f'(1)=0  f(x)=ax²+bln(x)

                                             f'(x)=2ax + b/x   2a+b=0  b=-2a

                                                                         f(x)=ax²-2alnx

                  ∫⁴1f(x)dx=27-8ln(4)  ∫⁴1 (ax² - 2aln(x))dx=27-8ln(4)=

             =[ax³/3 - 2a(x·lnx-x]⁴1= 64a/3 - 8aln(4) + 8´- a/3 + 2a=

                                                      27a - 8a·ln(4) a=1  b=-2

                        f(x) { 1+ a/(x-2)  si x<1     a+="" b/√x  =""  si="">1   >

                        f'(x) { -a/(x-2)²  si x<1         -b/2x√x    ="" si="">1       >

                                 f(1)=limx→1+f(x)= limx→1+(a + b/√x)=a+b

      limx→1-f(x)=limx→1-(1+ a/(x-2)=1-a         a+b=1-a    2a+b=1

                               f(1+)=limx→1+f'(x)=limx→1+(-b/(2x√x))=-b/2

           f(1-)=limx→1-f'(x)=limx→1-(a/(x-2)²)=-a      -a=-b/2   b=2a

                                                  2a+2a=4a=1   a=1/4   b=1/2

                 f(x)=(1-x²)e^-x  F(x)=I=∫(1-x²)e^-xdx    ∫udv=uv -∫vdz

                      u=1-x²   dv=e^-xdx   du=-2xdx    v=∫dv=∫e^-x=-e^x

            (1-x²)(-e^-x) -∫-e-x(-2xdx=-(1-x²)e^-x -2∫xe^-x=-(1-x²)e^-x-2I1

                            I1=∫xe^-xdx   u=x   dv=e^-xdx   du=dx   v=-e^-x

-x·e^-x -∫-e^-xdx= -x·e-x +∫e^-xdx=-x·e^-x -e^-x  F(x)=I=∫(1-x²)e^-xdx=

          -(1-x²)e^-x-2I1=-(1-x²)e^-x -2(-x·e^-x -e^-x)+k=e^-x(x² +2x+1) +k

                            F(-1)=0 e(1-2+1) +k=0    F(x)=e-x(x²+2x+1)

                        triángulos    A=1/2 xy  y² + x²=10²  y=√100-x²

       A(x)=½x √100-x²  A'(x)=½√100-x² + ½x ·[-2x/(2√100-x²)=

         =(50-x²)/(√100-x²)  A'(x)=0   50-x²=0  x=+√50 (isosceles)

                              y=√100-(√50)²=√50   A''(√50)=

                                                              f(x)=x²/4   g(x)=2√x

                                              g(x)  x=1  y=2     f(x)  x=1 y=1/4

                                              x²/4=2√x       x⁴-64x=x(x³-64)=0

                                                                             x=0   x=4

                                 ∫⁴o[(2√x)-(x²/4)]dx + ∫⁴o[(2x^1/2)-x²/4]dx =

                           [(2x^(1/2)+1)/(½+1)-(x³/12]⁴o=[(4√4³/3 -x³/12=

                                                        16√4/3 -64/12=5,33u²

            f(x) 1/x + lnx   f(x)=1/x+ln(x)   f'(x)=-1/x² +1/x   f'(x)=0

                    -1/x² + 1/x=0    x²=x    x²-x=0=x(x-1) x=0(no) x=1

                                      x=-1/e ,  x=1 ,  x=e  extrems absoluts

             f(1/e)=[1/(1/e)] + ln(1/e)= e + ln(1)-ln(e)= e - 1= 1,718

f(1)=1/1 + ln(1)=1+ ln(1)=1  f(e)= 1/e + ln(e)= (1/e) + 1 = 1,368

  e - 1 max abs(x=1/e)    1 min abs(x=1)    x=e   y-f(e)=f'(e)(x-e)

   f(x)= (1/x) + ln   f'(x)= -1/x² + 1/x     f(e)=(1/e) + ln(e)=2/e +1

  f'(e)=(-1/e²) + 1/e             y - (1/e + 1)=(-1/e²  +  1/e) · (x - e)

f(x) 2x²/(x+1)(x-2) x≠1,2  limx→1+ 2x²/(x+1)(x-2)=2/0-  x=-1AV

    limx→2+ 2x²/(x+1)(x-2)=8/0+  x=2AV  limx→+∞ 2x²/(x+1)(x-2)=

     limx→+∞ 2x²/x²=2   y=2AH   f'(x)=(-2x²-8x)/(x²-x-2)² x=0 x=-4

f'(-5)=<0 decr="">0><-4  >-4  >'(-2)=>0 crec -4<><0   >0   >'(1)=>0 dec x>0

                                                f(x)=2   x=-2    (-2,f(-2))= (-2,2)