Question

determine equação da reta tangente a função f(x) no ponto indicado f(x)=√x x=9

Answer 1

Olá

Equação da reta tangente

$\displaystyle\boxed{ \mathsf{y-f(x_0)=f'(x_0)\cdot (x-x_0)}}$


Dados

$\mathsf{f(x)= \sqrt{x} }\qquad\qquad\qquad\qquad \mathsf{x_0=9}$

Calculando f(x₀)

$\mathsf{f(9)= \sqrt{9} }\\\\\boxed{\mathsf{f(9)=3}}$


Derivando f(x)

$\displaystyle \mathsf{f(x)= \sqrt{x} }\\\\\mathsf{f(x)=x^{ \frac{1}{2} }}\\\\\mathsf{f'(x)= \frac{1}{2}x^{ \frac{1}{2}-1 } }\\\\\\\mathsf{f'(x)= \frac{1}{2}x^{- \frac{1}{2} } }\\\\\\\mathsf{f'(x)= \frac{1}{2 x^ \frac{1}{2} } }\\\\\\\mathsf{f'(x)= \frac{1}{2 \sqrt{x} } }$

Calculando f'(x₀)

$\displaystyle \mathsf{f'(9)= \frac{1}{2 \sqrt{9} } }\\\\\\\boxed{\mathsf{f'(9)= \frac{1}{6} }}$


Já temos todas as informações necessárias, então vamos substituir na fórmula

Relembrando os dados que temos

$\displaystyle \mathsf{f(x_0)=3}\qquad\qquad\qquad\qquad \mathsf{x_0=9}\\\\\mathsf{f'(x_0)= \frac{1}{6} }$

$\displaystyle\mathsf{y-f(x_0)=f'(x_0)\cdot (x-x_0)}\\\\\\\mathsf{y-3= \frac{1}{6}\cdot (x-9) }\\\\\\\mathsf{y-3= \frac{1}{6}x- \frac{3}{2} }\\\\\\\mathsf{y= \frac{1}{6} x- \frac{3}{2}+3 }\\\\\\\boxed{\boxed{\mathsf{y= \frac{1}{6}x+ \frac{3}{2} }}}\qquad\qquad\qquad\Longleftarrow\text{Equacao da reta tangente no ponto x}_0$