Question

find f such that $f(x)=\frac{4}{\sqrt{x}}$, $f(1)=19$.
$f(x)=\square$

Explanation:

Step1: Integrate \( f^{\prime}(x) \)

Given \( f^{\prime}(x)=\frac{4}{\sqrt{x}} = 4x^{-\frac{1}{2}} \).
Using the power - rule for integration \( \int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have:
\(\int f^{\prime}(x)dx=\int4x^{-\frac{1}{2}}dx\).
Since \(\int k\cdot g(x)dx=k\int g(x)dx\) (\(k = 4\) is a constant), then \(\int4x^{-\frac{1}{2}}dx=4\int x^{-\frac{1}{2}}dx\).
Applying the power - rule: \(4\times\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C=4\times\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C = 8\sqrt{x}+C\). So \(f(x)=8\sqrt{x}+C\).

Step2: Use the initial condition \( f(1) = 19 \)

Substitute \(x = 1\) and \(f(1)=19\) into \(f(x)=8\sqrt{x}+C\).
We get \(19=8\sqrt{1}+C\).
Since \(\sqrt{1}=1\), then \(19 = 8\times1+C\).
Solving for \(C\): \(C=19 - 8=11\).

Answer:

\(f(x)=8\sqrt{x}+11\)