Question

given the function $f(x)=sqrt{x^{3}}-\frac{sqrt{x}}{2}$, find $f(4)$. express your answer as a single fraction in simplest form.

Explanation:

Step1: Rewrite the function

Rewrite $f(x)=\sqrt{x^{3}}-\frac{\sqrt{x}}{2}$ as $f(x)=x^{\frac{3}{2}}-\frac{1}{2}x^{\frac{1}{2}}$.

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = x^{n}$, then $y^\prime=nx^{n - 1}$.
For $y = x^{\frac{3}{2}}$, $y^\prime=\frac{3}{2}x^{\frac{3}{2}-1}=\frac{3}{2}x^{\frac{1}{2}}$.
For $y=-\frac{1}{2}x^{\frac{1}{2}}$, $y^\prime=-\frac{1}{2}\times\frac{1}{2}x^{\frac{1}{2}-1}=-\frac{1}{4}x^{-\frac{1}{2}}$.
So, $f^\prime(x)=\frac{3}{2}x^{\frac{1}{2}}-\frac{1}{4}x^{-\frac{1}{2}}$.

Step3: Substitute $x = 4$

Substitute $x = 4$ into $f^\prime(x)$.
$f^\prime(4)=\frac{3}{2}\times\sqrt{4}-\frac{1}{4}\times\frac{1}{\sqrt{4}}$.
Since $\sqrt{4}=2$, we have $f^\prime(4)=\frac{3}{2}\times2-\frac{1}{4}\times\frac{1}{2}$.
$f^\prime(4)=3-\frac{1}{8}$.

Step4: Calculate the result

Find a common denominator: $3=\frac{24}{8}$.
So, $f^\prime(4)=\frac{24}{8}-\frac{1}{8}=\frac{24 - 1}{8}=\frac{23}{8}$.

Answer:

$\frac{23}{8}$