Question

Me ajudem por favor... como eu resolvo esse exercício e urgente !!

1º) calcule os declives dos gráficos das funções dadas, nos valores dados de x, e encontre as equações das retas tangentes e faça os gráficos.

a)$f(x)= x^{2} -2$ em  $x= \frac{3}{2}$, $x= \frac{-5}{3}$, $x= a$ qualquer.
b)$f(x)= \frac{2 x^{2} }{3} -3 x$ em $x=-1$, $x= \frac{2}{3}$, $x= \alpha$ qualquer.
c)$f(x)= \frac{1}{x}$ , em $x=1,x=2,x= \alpha \neq 0.$

2º)Encontre a reta normal á curva dada, passando pelo ponto de abscissa dada. e faça o gráfico.

a)$f(x)= x^{2}-2x+1$ em $x=0.$

3º) Encontre a reta tangente á curva $y= x^{2}$ com declive $m=-8$ e faça o gráfico.

Answer 1

Bom dia Luana!

Solução!

Para resolver esses exercícios,é necessario encontra o valor de y fazendo a substituição ainda con esse raciocinio: todas as funções precisam serem derivadas e substituir o valor de x para encontrar o coeficiente angular,feito isso podemos usar a formula da reta para estarmos explicitando a mesma.

Vamos chamar de m o coeficiente angular.


1º) calcule os declives dos gráficos das funções dadas, nos valores dados de x, e encontre as equações das retas tangentes e faça os gráficos.

$Exercicio ~~A\\\\\\ f(x)= x^{2} -2~~em ~~x= \frac{3}{2},~~x= \frac{-5}{3}\\\\\ y=( \frac{3}{2}) ^{2}-2\\\\\ y= \frac{9}{4}-2\\\\ y= \frac{1}{4} \\\\\ A( \frac{3}{2}, \frac{1}{4})\\\\\\ Calculo ~~do~~ coeficiente~~ angular.\\\\\ f(x)= x^{2} -2\\\\\  f'(x)= 2x\\\\\ m= 2.\frac{3}{2}\\\\\ m=3\Rightarrow~~coeficiente~~angular$

$equacao~~ da~~ reta.\\\\\\\ y-yA=m(x-xA)\\\\\ y- \frac{1}{4}=3(x- \frac{3}{2})\\\\\\\ y- \frac{1}{4}=3x- \frac{9}{2})\\\\ 4y-1=12x-18\\\\\ 4y=12x-18+1\\\\\ 4y=12x-17\\\\\ y= \frac{12x-17}{4} \Rightarrow~~Reta~~tangente$

$Exercicio ~~A\\\\\\ f(x)= x^{2} -2\\\\\\ y=(- \frac{5}{3})^{2}-2\\\\\ y= \frac{25}{9} -2\\\\\ y= \frac{7}{9} \\\\\ A( \frac{-5}{3}, \frac{7}{9})\\\\\ f(x)=x^{2} -2\\\\\ f'(x)= 2x\\\\\ m= 2.\frac{-5}{3}\\\\\ m= \frac{-10}{3} \Rightarrow coeficiente~~angular$

$y- \frac{7}{9}= \frac{-10}{3}(x+ \frac{5}{3})\\\\\ y- \frac{7}{9}= \frac{-10x}{3}- \frac{50}{9}\\\\\ 9y-7=-30x-50\\\\\ 9y=-30x-50+7\\\\\ 9y=-30x-43\\\\\ y= \frac{-30x-43}{9} \Rightarrow Reta~~tangente.$

$Exercicio~~b\\\\\\ f(x)= \frac{2 x^{2} }{3} -3x~~em~~x=-1~~,x= \frac{2}{3}\\\\\ y= \frac{2(-1)^{2} }{3} -3(-1)\\\\\ y= \frac{2}{3} +3 \\\\\ y= \frac{11}{3}\\\\\ B(-1, \frac{11}{3})\\\\\\ f(x)= \frac{2 x^{2} }{3} -3x\\\\\ f'(x)= \frac{4 x }{3} -3\\\\ m=\frac{4 .(-1) }{3} -3\\\\ m=\frac{-4 }{3} -3\\\\ m= \frac{-13}{3} \Rightarrow~~coeficiente~~angular$

$y- \frac{11}{3}=- \frac{13}{3}(x+1)\\\\\ y- \frac{11}{3}=- \frac{13x}{3}- \frac{13}{3} \\\\\ 3y-11=-13x-13\\\\\ 3y=-13x-13+11\\\\ 3y=-13x-2\\\\ y= \frac{-13x-2}{3} \Rightarrow Equacao~~ da~~ reta$

$Exercicio~~B\\\\\\ f(x)= \frac{2 x^{2} }{3} -3x\\\\ y=\frac{2 ( \frac{4}{9})^{2} }{3} -3( \frac{2}{3}) \\\\ y=\frac{2 ( \frac{4}{9})^}{3} -2 \\\\ y= \frac{8}{27}- 54\\\\ y= \frac{-46}{27} \\\\\ B( \frac{2}{3}, \frac{-46}{27})\\\\\\\ f(x)= \frac{2 x^{2} }{3} -3x\\\\\\ f'(x)= \frac{4 x }{3} -3\\\\\\ m= \frac{4 (\frac{2}{3} ) }{3} -3\\\\\\ m= \frac{8 }{9} -3\\\\\\ m= \frac{8-27}{9}\\\\\\ m= \frac{-19}{9} \Rightarrow coeficiente~~angular$

$y+ \frac{46}{27}= \frac{-19}{9}(x- \frac{2}{3})\\\\\ y+ \frac{46}{27}= \frac{-19x}{9} +\frac{38}{27}\\\\\ 27y+46=-57x+38\\\\\ 27y=-57x+38-46\\\\\ 27y=-57-8\\\\\ y= \frac{-57x-8}{27} \Rightarrow ~~equacao~~ da~~ reta.$


$Exercicio~~C\\\\\\ f(x)= \frac{1}{x} ~~em~~x=1~~e~~x=2\\\\ y=\frac{1}{1} \\\\ y=1\\\\\ C(1,1)\\\\ f(x)= \frac{1}{x}\\\\\\ f'(x)= x^{-1} \\\\\\ f'(x)= -x^{-1-1} \\\\\\ f'(x)= -x^{-2} \\\\\\ f'(x)= - \frac{1}{ x^{2} } \\\\\\ m= \frac{1}{1}\\\\ m=-1 \Rightarrow coeficiente~~angular$

$y-1=-1(x-1)\\\\ y-1=-x+1\\\\ y=-x+1+1\\\\ y=-x+2\Rightarrow equacao~~ da~~ reta.$


$Ecercicio~~C\\\\\ f(x)= \frac{1}{x}\\\\\ y= \frac{1}{2}\\\\\ C(2, \frac{1}{2})\\\\\ f'(x)= - \frac{1}{ x^{2} }\\\\ m= - \frac{1}{ 2^{2} }\\\\ m= - \frac{1}{ 4 }\Rightarrow coeficiente~~angular.$

$y- \frac{1}{2}= -\frac{1}{4}(x-2)\\\\\ y- \frac{1}{2}= -\frac{1x}{4}+ \frac{1}{2} \\\\\ 4y-2=-x+2\\\\\ 4y=-x+2+2\\\\\ 4y=-x+4\\\\ y= \frac{-x}{4}+1 \Rightarrow equacao~~da~~reta$


2º)Encontre a reta normal á curva dada, passando pelo ponto de abscissa dada. e faça o gráfico.

Observação: reta normal tem que ser perpendicular a reta tangente.

$Exercicio~~A\\\\\ f(x)= x^{2} -2x+1~~em~~x=0\\\\ y=0^{2} -2.0+1\\\\ y=1\\\\\ A(0,1)\\\\\ f(x)= x^{2} -2x+1\\\\\ f'(x)= 2x -2\\\\\ m=2.0-2\\\\ m=-2$

$y-1=-2(x-0)\\\\\ y=-2x+1\Rightarrow equacao~~da ~~reta~~tangente.$

Para determinar a reta tangente basta inverter o coeficiente angular.
$m= \frac{1}{2}\\\\\ y-1= \frac{1}{2}(x-0)\\\\ y-1= \frac{x}{2}\\\\\ 2y-2= x\\\\\ 2y=x+2\\\\\ y= \frac{x}{2}+1\Rightarrow equacao~~da~~reta~~normal~~em~~vermelho.$

Exercicio   3

$y= x^{2} \\\\\ m=-8\\\\\ m=2x\\\\ 2x=-8\\\\ x= \frac{-8}{2}\\\\\ x=-4\\\\\ substituindo!\\\\\ y= x^{2} \\\\ y=(-4)^{2}\\\\ y=16\\\\\ D(-4,16)\\\\\ y-16=-8(x+4)\\\\ y-16=-8x-32\\\\ y=-8x-32+16\\\\\ y=-8x-16 \Rightarrow equacao~~da~~reta$

Os gráficos estão anexo.

Bom dia!
Bons estudos!