Question

find f such that ( f^{prime}(x)=\frac{7}{sqrt{x}} ), ( f(16)=70 ).
( f(x)=square )

Explanation:

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

Given \( f^{\prime}(x)=\frac{7}{\sqrt{x}} = 7x^{-\frac{1}{2}}\).
Using the power - rule for integration \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have:
\(\int f^{\prime}(x)dx=\int7x^{-\frac{1}{2}}dx\).
Since \(\int k\cdot g(x)dx=k\int g(x)dx\) (\(k = 7\) is a constant), then \(\int7x^{-\frac{1}{2}}dx=7\int x^{-\frac{1}{2}}dx\).
Applying the power - rule: \(7\times\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C=7\times\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C = 14\sqrt{x}+C\). So \(f(x)=14\sqrt{x}+C\).

Step2: Use the initial condition \(f(16) = 70\)

Substitute \(x = 16\) and \(f(16)=70\) into \(f(x)=14\sqrt{x}+C\).
We get \(70=14\sqrt{16}+C\).
Since \(\sqrt{16}=4\), then \(70=14\times4+C\).
\(70 = 56+C\).
Solve for \(C\): \(C=70 - 56=14\).

Answer:

\(f(x)=14\sqrt{x}+14\)