Question

find f such that f(x)=\frac{5}{\sqrt{x}}, f(1)=18. f(x)=\square

Explanation:

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

Given \( f^{\prime}(x)=\frac{5}{\sqrt{x}} = 5x^{-\frac{1}{2}} \).
Using the power - rule for integration \( \int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have:
\(\int f^{\prime}(x)dx=\int5x^{-\frac{1}{2}}dx\).
Since \(\int k\cdot g(x)dx=k\int g(x)dx\) (\(k = 5\) is a constant), then \(\int5x^{-\frac{1}{2}}dx=5\int x^{-\frac{1}{2}}dx\).
Applying the power - rule: \(5\times\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C=5\times\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C = 10\sqrt{x}+C\). So \(f(x)=10\sqrt{x}+C\).

Step2: Use the initial condition \( f(1) = 18 \)

Substitute \(x = 1\) and \(f(1)=18\) into \(f(x)=10\sqrt{x}+C\).
We get \(18=10\sqrt{1}+C\).
Since \(\sqrt{1}=1\), then \(18 = 10\times1+C\).
Solving for \(C\): \(C=18 - 10=8\).

Answer:

\(f(x)=10\sqrt{x}+8\)