Question

9 $f(x)=\frac {2}{1+x^{2}}$ $f(-1)=3$.

to advance in the circuit, locate $f(\sqrt {3})$.

Explanation:

Step1: Integrate \( f'(x) \)

We know that \(\int\frac{1}{1 + x^{2}}dx=\arctan(x)+C\). So, \(\int f'(x)dx=\int\frac{2}{1 + x^{2}}dx\). Using the constant - multiple rule of integration \(\int k\cdot g(x)dx=k\int g(x)dx\) (\(k = 2\) and \(g(x)=\frac{1}{1 + x^{2}}\)), we get \(f(x)=2\arctan(x)+C\).

Step2: Find the value of \(C\)

Since \(f(-1) = 3\), substitute \(x=-1\) into \(f(x)=2\arctan(x)+C\). We know that \(\arctan(-1)=-\frac{\pi}{4}\). So, \(3=2\times(-\frac{\pi}{4})+C\). Then \(C = 3+\frac{\pi}{2}\).

Step3: Calculate \(f(\sqrt{3})\)

Substitute \(x = \sqrt{3}\) into \(f(x)=2\arctan(x)+3+\frac{\pi}{2}\). Since \(\arctan(\sqrt{3})=\frac{\pi}{3}\), then \(f(\sqrt{3})=2\times\frac{\pi}{3}+3+\frac{\pi}{2}\).

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Answer:

\(f(\sqrt{3})=\frac{7\pi}{6}+3\)