Question

homework3: problem 24
(1 point)
results for this submission
2 of the answers are not correct.
let
$f(x)=7x^{1/5}(x^{3}-2)$.
evaluate the following specific values of $f$:
(a) $f(3)=174.28$
(b) $f(6)=323.21$
note: you can earn partial credit on this problem.
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Explanation:

Step1: Expand the function

First, expand \(f(x)=7x^{1/5}(x^{3}-2)=7x^{1/5 + 3}-14x^{1/5}=7x^{16/5}-14x^{1/5}\).

Step2: Differentiate using the power rule

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

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

Substitute \(x = 3\) into \(f^\prime(x)\):
\(f^\prime(3)=\frac{112}{5}\times3^{\frac{11}{5}}-\frac{14}{5}\times3^{-\frac{4}{5}}\).
\(3^{\frac{11}{5}}=3^{2+\frac{1}{5}} = 9\times3^{\frac{1}{5}}\approx9\times1.2457\), \(3^{-\frac{4}{5}}=\frac{1}{3^{\frac{4}{5}}}\approx\frac{1}{4.3267}\).
\(f^\prime(3)=\frac{112}{5}\times9\times1.2457-\frac{14}{5}\times\frac{1}{4.3267}\approx\frac{112\times9\times1.2457}{5}-\frac{14}{5\times4.3267}\approx252.99 - 0.65\approx252.34\).

Step4: Calculate \(f^\prime(6)\)

Substitute \(x = 6\) into \(f^\prime(x)\):
\(f^\prime(6)=\frac{112}{5}\times6^{\frac{11}{5}}-\frac{14}{5}\times6^{-\frac{4}{5}}\).
\(6^{\frac{11}{5}}=6^{2+\frac{1}{5}}=36\times6^{\frac{1}{5}}\approx36\times1.4307\), \(6^{-\frac{4}{5}}=\frac{1}{6^{\frac{4}{5}}}\approx\frac{1}{9.506}\).
\(f^\prime(6)=\frac{112}{5}\times36\times1.4307-\frac{14}{5\times9.506}\approx\frac{112\times36\times1.4307}{5}-\frac{14}{47.53}\approx1153.77-0.29\approx1153.48\).

Answer:

(A) \(f^\prime(3)\approx252.34\)
(B) \(f^\prime(6)\approx1153.48\)