question solve the following initial - value ...

# Explanation: ## Step1: Integrate $f'(x)$ Integrate $\cos(x)-\sin(x)$ with respect to $x$. We know that $\int\cos(x)dx=\sin(x)+C_1$ and $\int\sin(x)dx = -\cos(x)+C_2$. So, $f(x)=\int(\cos(x)-\sin(x))dx=\sin(x)+\cos(x)+C$. ## Step2: Use the initial - condition Given $f(\frac{\pi}{3})=\frac{\sqrt{3}}{2}+\frac{7}{2}$. Substitute $x = \frac{\pi}{3}$ into $f(x)=\sin(x)+\cos(x)+C$. We have $\sin(\frac{\pi}{3})+\cos(\frac{\pi}{3})+C=\frac{\sqrt{3}}{2}+\frac{7}{2}$. Since $\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$ and $\cos(\frac{\pi}{3})=\frac{1}{2}$, then $\frac{\sqrt{3}}{2}+\frac{1}{2}+C=\frac{\sqrt{3}}{2}+\frac{7}{2}$. ## Step3: Solve for $C$ Subtract $\frac{\sqrt{3}}{2}+\frac{1}{2}$ from both sides of the equation $\frac{\sqrt{3}}{2}+\frac{1}{2}+C=\frac{\sqrt{3}}{2}+\frac{7}{2}$. We get $C = 3$. # Answer: $\sin(x)+\cos(x)+3$