Question

solve the following initial - value problem. $f(x)=-2sin(x)$, $f(\frac{pi}{4})=sqrt{2}+7$. 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 $\int\sin(x)dx=-\cos(x)+C$, we have $f(x)=\int - 2\sin(x)dx = 2\cos(x)+C$.

Step2: Use the initial - condition

Given $f(\frac{\pi}{4})=\sqrt{2}+7$. Substitute $x = \frac{\pi}{4}$ into $f(x)=2\cos(x)+C$. So $f(\frac{\pi}{4})=2\cos(\frac{\pi}{4})+C$. Since $\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$, we get $2\times\frac{\sqrt{2}}{2}+C=\sqrt{2}+7$.

Step3: Solve for $C$

Simplify the left - hand side of the equation $2\times\frac{\sqrt{2}}{2}+C=\sqrt{2}+7$, which is $\sqrt{2}+C=\sqrt{2}+7$. Subtract $\sqrt{2}$ from both sides, we find that $C = 7$.

Answer:

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