let (f(x)=7x^{2/5}-4x^{3/7}). evaluate each of the following: 1. (f(3)=) 2. (f(5)=)

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Accepted Answer

# Explanation:
## Step1: Find the derivative of $f(x)$
Use the power - rule $(x^n)^\prime=nx^{n - 1}$.
$$
\begin{align*}
f(x)&=7x^{\frac{2}{5}}-4x^{\frac{3}{7}}\
f^\prime(x)&=7\times\frac{2}{5}x^{\frac{2}{5}-1}-4\times\frac{3}{7}x^{\frac{3}{7}-1}\
&=\frac{14}{5}x^{-\frac{3}{5}}-\frac{12}{7}x^{-\frac{4}{7}}\
&=\frac{14}{5x^{\frac{3}{5}}}-\frac{12}{7x^{\frac{4}{7}}}
\end{align*}
$$
## Step2: Evaluate $f^\prime(3)$
Substitute $x = 3$ into $f^\prime(x)$.
$$
\begin{align*}
f^\prime(3)&=\frac{14}{5\times3^{\frac{3}{5}}}-\frac{12}{7\times3^{\frac{4}{7}}}\
&=\frac{14}{5\times\sqrt[5]{27}}-\frac{12}{7\times\sqrt[7]{81}}
\end{align*}
$$
$$
\begin{align*}
\frac{14}{5\times\sqrt[5]{27}}&\approx\frac{14}{5\times2.908}&\text{(since }\sqrt[5]{27}\approx2.908\text{)}\
&=\frac{14}{14.54}\approx0.963\
\frac{12}{7\times\sqrt[7]{81}}&\approx\frac{12}{7\times2.167}&\text{(since }\sqrt[7]{81}\approx2.167\text{)}\
&=\frac{12}{15.169}\approx0.791\
f^\prime(3)&\approx0.963 - 0.791=0.172
\end{align*}
$$
## Step3: Evaluate $f^\prime(5)$
Substitute $x = 5$ into $f^\prime(x)$.
$$
\begin{align*}
f^\prime(5)&=\frac{14}{5\times5^{\frac{3}{5}}}-\frac{12}{7\times5^{\frac{4}{7}}}\
&=\frac{14}{5\times\sqrt[5]{125}}-\frac{12}{7\times\sqrt[7]{625}}\
&=\frac{14}{5\times5^{\frac{3}{5}}}-\frac{12}{7\times5^{\frac{4}{7}}}\
&=\frac{14}{5\times3.017}-\frac{12}{7\times2.924}\
&=\frac{14}{15.085}-\frac{12}{20.468}\
&\approx0.928 - 0.586 = 0.342
\end{align*}
$$
# Answer:
1. $f^\prime(3)\approx0.172$
2. $f^\prime(5)\approx0.342$

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Question

Explanation:

Step1: Find the derivative of \(f(x)\)

Step2: Evaluate \(f^\prime(3)\)

Step3: Evaluate \(f^\prime(5)\)

Answer: