Question

assignment 4: problem 5 (1 point) let $f(x)=\frac{5x^{2}+6x + 5}{sqrt{x}}$. find the following: $f(x)=$ $f(4)=$

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\frac{5x^{2}+6x + 5}{\sqrt{x}}$ as $f(x)=5x^{\frac{3}{2}}+6x^{\frac{1}{2}}+5x^{-\frac{1}{2}}$ using the rule $\frac{x^{m}}{x^{n}}=x^{m - n}$.

Step2: Differentiate term - by - term

Using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have:
$\frac{d}{dx}(5x^{\frac{3}{2}})=5\times\frac{3}{2}x^{\frac{3}{2}-1}=\frac{15}{2}x^{\frac{1}{2}}$,
$\frac{d}{dx}(6x^{\frac{1}{2}})=6\times\frac{1}{2}x^{\frac{1}{2}-1}=3x^{-\frac{1}{2}}$,
$\frac{d}{dx}(5x^{-\frac{1}{2}})=5\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=-\frac{5}{2}x^{-\frac{3}{2}}$.
So, $f^{\prime}(x)=\frac{15}{2}x^{\frac{1}{2}}+3x^{-\frac{1}{2}}-\frac{5}{2}x^{-\frac{3}{2}}$.

Step3: Evaluate $f^{\prime}(4)$

Substitute $x = 4$ into $f^{\prime}(x)$:
$f^{\prime}(4)=\frac{15}{2}\times\sqrt{4}+3\times\frac{1}{\sqrt{4}}-\frac{5}{2}\times\frac{1}{\sqrt{4^{3}}}$
$=\frac{15}{2}\times2 + 3\times\frac{1}{2}-\frac{5}{2}\times\frac{1}{8}$
$=15+\frac{3}{2}-\frac{5}{16}$
$=\frac{240 + 24-5}{16}=\frac{259}{16}$.

Answer:

$f^{\prime}(x)=\frac{15}{2}x^{\frac{1}{2}}+3x^{-\frac{1}{2}}-\frac{5}{2}x^{-\frac{3}{2}}$
$f^{\prime}(4)=\frac{259}{16}$