Question

Resolução de aplicação de derivadas

Answer 1

$f(x)=\,^{3}\!\!\!\sqrt{x}\\ \\ f(x)=x^{1/3}\\ \\ \\ f'(x)=\frac{1}{3}\,x^{\frac{1}{3}-1}\\ \\ f'(x)=\frac{1}{3}\,x^{\frac{1-3}{3}}\\ \\ f'(x)=\frac{1}{3}\,x^{-2/3}\\ \\ f'(x)=\frac{1}{3x^{2/3}}\\ \\ f'(x)=\frac{1}{3\,^{3}\!\!\!\sqrt{x^{2}}}$


Pelo Teorema do Valor Médio, existe um $c$

$0<c<8,$

tal que


$f'(c)=\frac{f(8)-f(0)}{8-0}\\ \\ \frac{1}{3\,^{3}\!\!\!\sqrt{c^{2}}}=\frac{\,^{3}\!\!\!\sqrt{8}-\,^{3}\!\!\!\sqrt{0}}{8-0}\\ \\ \frac{1}{3\,^{3}\!\!\!\sqrt{c^{2}}}=\frac{2-0}{8}\\ \\ \frac{1}{3\,^{3}\!\!\!\sqrt{c^{2}}}=\frac{1}{4}\\ \\ 3\,^{3}\!\!\!\sqrt{c^{2}}=4\\ \\ ^{3}\!\!\!\sqrt{c^{2}}=\frac{4}{3}\\ \\ c^{2}=\left(\frac{4}{3} \right )^{3}\\ \\ c^{2}=\frac{64}{27}\\ \\ c=\sqrt{\frac{64}{27}}\\ \\ \boxed{\begin{array}{c}c=\frac{8}{3\sqrt{3}} \end{array}}$