Question

assignment 4: problem 7 (1 point) if $f(x)=3sqrt{x}(x^{3}-3sqrt{x}+6)$, find $f(16)$. $f(16)=square$ you have attempted this problem 0 times. you have unlimited attempts remaining.

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=3\sqrt{x}(x^{3}-3\sqrt{x}+6)$ as $f(x)=3x^{\frac{1}{2}}(x^{3}-3x^{\frac{1}{2}} + 6)=3x^{\frac{7}{2}}-9x + 18x^{\frac{1}{2}}$.

Step2: Differentiate the function

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $f'(x)=3\times\frac{7}{2}x^{\frac{5}{2}}-9+18\times\frac{1}{2}x^{-\frac{1}{2}}=\frac{21}{2}x^{\frac{5}{2}}-9 + 9x^{-\frac{1}{2}}$.

Step3: Substitute $x = 16$

Substitute $x = 16$ into $f'(x)$. Since $\sqrt{16}=4$, then $x^{\frac{5}{2}}=(x^{\frac{1}{2}})^{5}=4^{5}=1024$ and $x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}=\frac{1}{4}$.
$f'(16)=\frac{21}{2}\times1024-9 + 9\times\frac{1}{4}$.
First, $\frac{21}{2}\times1024=21\times512 = 10752$.
Then $f'(16)=10752-9+\frac{9}{4}=10743+\frac{9}{4}=\frac{42972 + 9}{4}=\frac{42981}{4}=10745.25$.

Answer:

$10745.25$