Question

solve the following initial - value problem. $f^{prime}(x)=-2sin(x)$, $f(\frac{pi}{4})=sqrt{2}+7$. provide your answer below: $f(x)=square$

Explanation:

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

We know that if \(f'(x)=- 2\sin(x)\), then by the integral formula \(\int\sin(x)dx=-\cos(x)+C\), \(f(x)=\int - 2\sin(x)dx = 2\cos(x)+C\).

Step2: Use the initial - condition

Given \(f(\frac{\pi}{4})=\sqrt{2}+7\). Substitute \(x = \frac{\pi}{4}\) into \(f(x)=2\cos(x)+C\). We have \(f(\frac{\pi}{4})=2\cos(\frac{\pi}{4})+C\). Since \(\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\), then \(2\times\frac{\sqrt{2}}{2}+C=\sqrt{2}+7\). Simplifying the left - hand side gives \(\sqrt{2}+C=\sqrt{2}+7\). Solving for \(C\), we get \(C = 7\).

Step3: Write the final function

So \(f(x)=2\cos(x)+7\).

Answer:

\(f(x)=2\cos(x)+7\)