Question

if $f(x)=\frac{sqrt{x}-4}{sqrt{x}+4}$, find:
$f(x)=$
$f(4)=$
question help: video message instructor
submit question jump to answer

Explanation:

Step1: Apply the quotient rule

The quotient rule states that if \(y = \frac{u}{v}\), then \(y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Let \(u=\sqrt{x}-4=x^{\frac{1}{2}} - 4\) and \(v=\sqrt{x}+4=x^{\frac{1}{2}}+4\).

First, find \(u^\prime\) and \(v^\prime\). Using the power rule \((x^{n})^\prime=nx^{n - 1}\), we have \(u^\prime=\frac{1}{2}x^{-\frac{1}{2}}\) and \(v^\prime=\frac{1}{2}x^{-\frac{1}{2}}\).

Step2: Substitute into the quotient rule formula

$$
LATEXBLOCK0
$$

Step3: Evaluate \(f^\prime(4)\)

Substitute \(x = 4\) into \(f^\prime(x)\). Since \(x^{\frac{1}{2}}=\sqrt{4} = 2\) and \(\sqrt{x}+4=2 + 4=6\)

$$ f^\prime(4)=\frac{4}{2\times6^{2}}=\frac{4}{2\times36}=\frac{1}{18} $$

Answer:

\(f^\prime(x)=\frac{4}{x^{\frac{1}{2}}(\sqrt{x}+4)^{2}}\)

\(f^\prime(4)=\frac{1}{18}\)