Question

dada a função f (x)= 0,50 x3 + 4,0 determine a equação da reta tangente ao grafico da função no ponto x=3

Answer 1

Olá

Equação da reta tangente:

$\mathsf{RT:\qquad y-f(x_0)=f'(x_0)\cdot (x-x_0)}\\\\\\\mathsf{x_0=3\qquad\qquad \Longleftarrow \qquad conforme~o~enunciado}$


Encontrando f(x) no ponto x=3

$\mathsf{f(x)=0,5x^3+4}\\\\\mathsf{f(x_0)=f(3)=0,5\cdot(3)^3+4}\\\\\boxed{\mathsf{f(3)=17,5}}$



Derivando f(x)


$\mathsf{f(x)=0,5x^3+4}\\\\\\\mathsf{f'(x)=3\cdot 0,5x^{3-1}+0}\\\\\\\boxed{\mathsf{f'(x)=1,5x^2}}$


Encontrando f'(x) no ponto x=3

$\mathsf{f'(x)=1,5x^2}\\\\\mathsf{f'(x_0)=f'(3)=1,5\cdot (3)^2}\\\\\\\boxed{\mathsf{f'(3)=13,5}}$



Todos os dados que temos até agora

$\mathsf{x_0=3}\\\\\mathsf{f(x_0)=17,5}\\\\\mathsf{f'(x_0)=13,5}\\\\\\\\\text{Substituindo na formula}\\\\\\\mathsf{RT:\qquad y-f(x_0)=f'(x_0)\cdot (x-x_0)}\\\\\\\mathsf{\qquad\qquad y-17,5=(13,5)\cdot (x-3)}\\\\\\\mathsf{\qquad\qquad y-17,5=13,5x-40,5}\\\\\\\mathsf{\qquad\qquad y=13,5x-40,5+17,5}\\\\\\\boxed{\boxed{\mathsf{RT:\qquad y=13,5x-23}}}\qquad\qquad\Longleftarrow\qquad\text{Reta tang. em }\mathsf{x_0=3}$




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