Question

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

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. \(f^\prime(5)\approx0.342\)