Question

POR FAVOR ME AJUDEM!! É URGENTE !!!
Determine o coeficiente angular do gráfico da função em um ponto dado.Em seguida, determine uma equação para a reta tangente ao gráfico naquele ponto.

a) f(x) = x^2 +x (2,5)
b) f(x) = x- 2x^2 (1,-1)
c) g(x) = x/x-2 (3,3)
d) g(x) = 8/x^2 (2,2)

Answer 1

coeficiente angular da função num determinado ponto é a sua derivada:
$\displaystyle a)\frac{d}{dx}x^2+x=2x+1\\\\b)\frac{d}{dx}x-2x^2=1-4x\\\\c)\frac{d}{dx}\frac{x}{x-2}=\frac{(x-2)-(x)}{(x-2)^2}=\frac{-2}{(x-2)^2}\\\\d)\frac{d}{dx}\frac{8}{x^2}=\frac{-2x.8}{x^4}=-\frac{16x}{x^4}$
as equações das retas tangentes:
$y=mx-mx_{_0}+y_{_0}$
$\displaystyle a) (2,\ 5)\\y=(2x+1)x-(2x+1)2+5=2x^2+x-4x+7=2x^2-3x+7\\\\b)\ (1,\ -1)\\y=(1-4x)x-(1-4x)1+(-1)=x-4x^4-1+4x-1\implies \\y=4x^4+5x-2\\\\\displaystylec)\ (3,\ 3)\\ y=-\frac{2}{(x-2)^2}x+\frac{2}{(x-2)^2}3+3\implies \\\\-\frac{2x}{(x^2-4x+4)}+\frac{6}{x^2-4x+4}+3=\frac{6-2x}{(x-2)^2}+3\implies\\\\\frac{6-2x}{(x-2)^2}+\frac{3(x-2)^2}{(x-2)^2}=\frac{3(x-2)^2+6-2x}{(x-2)^2}$$\displaystyled)\ (2,\ 2)\\y=-\frac{16x}{x^4}x-(-\frac{16x}{x^4})2+2\implies y=\frac{16x^2}{x^4}+\frac{32x}{2x^4}+2\implies\\\\ y=\frac{16}{x^2}+\frac{32}{2x^3}+2\implies y=\frac{16.2x^3}{2x^3x^2}+\frac{32x^2}{x^22x^3}+2\implies\\\\y=\frac{32x^3+32x^2}{2x^5}+2\implies y=\frac{32x^3+32x^2}{2x^5}+\frac{2.2x^5}{2x^5}\implies\\\\y=\frac{32x^3+32x^2+4x^5}{2x^5}=\frac{x^2(32x+32+4x^3)}{2x^5}\implies \\\\y=\frac{4x^3+32x+32}{2x^3}$