Question

determine a equação da recta tangente a curva f(x)=√x no ponto x=4​

Answer 1

Equação da reta tangente

$\boxed{\boxed{\boxed{\boxed{\mathsf{y=y_{0}+f'(x_{0})(x-x_{0})}}}}}$

$\dotfill$

$\mathsf{f(x)=\sqrt{x}}\\\mathsf{f(4)=\sqrt{4}=2}\\\mathsf{P(4,2)}$

$\mathsf{f'(x)=\dfrac{1}{2\sqrt{x}}}\\\mathsf{f'(4)=\dfrac{1}{2\sqrt{4}}=\dfrac{1}{2.2}=\dfrac{1}{4}}$

$\mathsf{y=2+\dfrac{1}{4}(x-2)}\\\mathsf{y=2+\dfrac{1}{4}x-\dfrac{1}{2}}\\\huge\boxed{\boxed{\boxed{\boxed{\mathsf{y=\dfrac{1}{4}x+\dfrac{3}{2}}}}}}$