encontre a equação da reta tangente de curva a, plote o gráfico da função e a reta obtida, . f(x)= (x^2-2x+1)∙ 3^x; no ponto de abscissa x=-2
💡 2 Respostas
Eduardo Germano
Ponto de tangencia (-1,1/7)
f^' (x)=d/dx ((2x+1)/(3x-4))
f^' (x)=(d/dx (2x+1)*(3x-4)-(2x+1)*d/dx (3x-4))/(3x-4)^2
f^' (x)=(2(3x-4)-(2x+1)x3)/(3x-4)^2
f^' (x)= -11/(3x-4)^2
f^' (x)= -11/(3(-1)-4)^2 =-11/49
f(x)=(2x+1)/(3x-4),x=-1:m= -11/49
〖A reta com inclinação m=-11/49 que passa por (-1,1/7): f〗^' (x)=-11/49 x-4/49
f(x)=-11/49 x-4/49
Ponto de tangencia (-1,1/7)
f^' (x)=d/dx ((2x+1)/(3x-4))
f^' (x)=(d/dx (2x+1)*(3x-4)-(2x+1)*d/dx (3x-4))/(3x-4)^2
f^' (x)=(2(3x-4)-(2x+1)x3)/(3x-4)^2
f^' (x)= -11/(3x-4)^2
f^' (x)= -11/(3(-1)-4)^2 =-11/49
f(x)=(2x+1)/(3x-4),x=-1:m= -11/49
〖A reta com inclinação m=-11/49 que passa por (-1,1/7): f〗^' (x)=-11/49 x-4/49
f(x)=-11/49 x-4/49
PARTE 2
Ponto de tangencia (-2,1)
f^' (x)=d/dx ((x^2-2x+1)*3^x )
f^' (x)=d/dx(x^2*3^x-2x*3^x+3^x)
f^' (x)=d/dx (x^2*3^x )+d/dx (-2x*3^x )+d/dx(3^x)
f^' (x)=2x*3^x+x^2*ln(3)*3^x+d/dx (-2x*3^x )+d/dx(3^x)
f^' (x)=2x*3^x+x^2*ln(3)*3^x-2*3^x-2x*ln(3)*3^x+d/dx(3^x)
f^' (x)=2x*3^x+x^2*ln(3)*3^x-2*3^x-2x*ln(3)*3^3+ln(3)*3^x
f^' (x)=2x*3^x+ln(3) x^2*3^x-2ln(3)x*3^x+ln(3)*3^x
f^' (x)=2x*3^x+ln(3) x^2*3^x-2*3^x-2ln(3)x*3^x+ln(3)*3^x
f(x)=(x^2-2x+1) 3^x,x=-2:m=ln(3)-2/3
A reta com inclinação m=ln(3)-2/3 que passa por (-2,1):f(x)=(ln(3)-2/3)
f(x)=(ln(3)-2/3)x+2ln(3)-1/3
Thalles Araujo
Ponto de tangencia (-1,1/7)
f^' (x)=d/dx ((2x+1)/(3x-4))
f^' (x)=(d/dx (2x+1)*(3x-4)-(2x+1)*d/dx (3x-4))/(3x-4)^2
f^' (x)=(2(3x-4)-(2x+1)x3)/(3x-4)^2
f^' (x)= -11/(3x-4)^2
f^' (x)= -11/(3(-1)-4)^2 =-11/49
f(x)=(2x+1)/(3x-4),x=-1:m= -11/49
〖A reta com inclinação m=-11/49 que passa por (-1,1/7): f〗^' (x)=-11/49 x-4/49
f(x)=-11/49 x-4/49
Ponto de tangencia (-1,1/7)
f^' (x)=d/dx ((2x+1)/(3x-4))
f^' (x)=(d/dx (2x+1)*(3x-4)-(2x+1)*d/dx (3x-4))/(3x-4)^2
f^' (x)=(2(3x-4)-(2x+1)x3)/(3x-4)^2
f^' (x)= -11/(3x-4)^2
f^' (x)= -11/(3(-1)-4)^2 =-11/49
f(x)=(2x+1)/(3x-4),x=-1:m= -11/49
〖A reta com inclinação m=-11/49 que passa por (-1,1/7): f〗^' (x)=-11/49 x-4/49
f(x)=-11/49 x-4/49
PARTE 2
Ponto de tangencia (-2,1)
f^' (x)=d/dx ((x^2-2x+1)*3^x )
f^' (x)=d/dx(x^2*3^x-2x*3^x+3^x)
f^' (x)=d/dx (x^2*3^x )+d/dx (-2x*3^x )+d/dx(3^x)
f^' (x)=2x*3^x+x^2*ln(3)*3^x+d/dx (-2x*3^x )+d/dx(3^x)
f^' (x)=2x*3^x+x^2*ln(3)*3^x-2*3^x-2x*ln(3)*3^x+d/dx(3^x)
f^' (x)=2x*3^x+x^2*ln(3)*3^x-2*3^x-2x*ln(3)*3^3+ln(3)*3^x
f^' (x)=2x*3^x+ln(3) x^2*3^x-2ln(3)x*3^x+ln(3)*3^x
f^' (x)=2x*3^x+ln(3) x^2*3^x-2*3^x-2ln(3)x*3^x+ln(3)*3^x
f(x)=(x^2-2x+1) 3^x,x=-2:m=ln(3)-2/3
A reta com inclinação m=ln(3)-2/3 que passa por (-2,1):f(x)=(ln(3)-2/3)
f(x)=(ln(3)-2/3)x+2ln(3)-1/3
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