let\n\n$f(x)=7x^{1/5}(x^{3}-2)$.\n\nevaluate the following specific values of $f$:\n(a) $f(3)=$ 246.13\n(b)…

Explanation:

Step1: Expand the function

$$ \begin{align*} f(x)&=7x^{1/5}(x^{3}-2)\ &=7x^{1/5 + 3}-14x^{1/5}\ &=7x^{16/5}-14x^{1/5} \end{align*} $$

Step2: Differentiate using power rule

The power rule is ((x^n)^\prime=nx^{n - 1}). For (y = 7x^{16/5}-14x^{1/5}), then (f^\prime(x)=7\times\frac{16}{5}x^{\frac{16}{5}-1}-14\times\frac{1}{5}x^{\frac{1}{5}-1}) (f^\prime(x)=\frac{112}{5}x^{\frac{11}{5}}-\frac{14}{5}x^{-\frac{4}{5}})

Step3: Calculate (f^\prime(3))

Substitute (x = 3) into (f^\prime(x)): [ \begin{align*} f^\prime(3)&=\frac{112}{5}\times3^{\frac{11}{5}}-\frac{14}{5}\times3^{-\frac{4}{5}}\ &=\frac{112}{5}\times3^{2+\frac{1}{5}}-\frac{14}{5}\times3^{- \frac{4}{5}}\ &=\frac{112}{5}\times9\times3^{\frac{1}{5}}-\frac{14}{5}\times\frac{1}{3^{\frac{4}{5}}}\ &\approx\frac{112}{5}\times9\times1.2457-\frac{14}{5}\times\frac{1}{2.2795}\ &=\frac{112\times9\times1.2457}{5}-\frac{14}{5\times2.2795}\ &=\frac{112\times11.2113}{5}-\frac{14}{11.3975}\ &=\frac{1255.6656}{5}-1.2283\ & = 251.1331-1.2283\ &=249.90 \end{align*} ]

Answer:

(f^\prime(3)\approx249.90)