Question

Encontre a equação da reta tangente à curva no ponto dado: (a) f(x) = x +√x (1, 2) (b) f(x) = x^2 + 2e^x (0, 2)

Answer 1

Resposta:

a) $3x-2y+1=0$

$b)2x-y+2=0$

Explicação passo a passo:

Encontre a equação da reta tangente à curva no ponto dado:

(a) f(x) = x +√x  ;   (1, 2)

(b) f(x) = x^2 + 2e^x  ;   (0, 2)

a) Cálculo do coeficiente angular:

$f'(x) =1+\frac{x'}{2\sqrt{x} } \\\\f'(1)=1+\frac{1}{2\sqrt{1} } \\\\f'(1)=1+\frac{1}{2.1} \\\\f'(x)=1+\frac{1}{2} \\\\f'(x)=\frac{3}{2}$

Equação da reta

Seja P(1,2)

$y-y_P=f'(1)(x-x_P)\\\\y-2=\frac{3}{2} (x-1)\\\\2y-4=3x-3\\\\2y-3x-4+3=0\\\\-3x+2y-1=0\\\\3x-2y+1=0$

$b)f(x)=x^2+2e^x\\\\f'(x)=2x+2e^x\\\\f'(0)=2.0+2.e^0\\\\f'(0)=0+2.1\\\\f'(0)=2\\\\y-2=2(x-0)\\\\y-2=2x\\\\-2x+y-2=0\\\\2x-y+2=0$