the graph of which function passes through (0,4) and has a minimum value at $\\left(\\frac{3\\pi}{2},3\\right…

Explanation:

Step1: Check point (0,4) for each function

• For $f(x)=\sin(x)+4$, when $x = 0$, $f(0)=\sin(0)+4=0 + 4=4$.
• For $f(x)=\cos(x)+3$, when $x = 0$, $f(0)=\cos(0)+3=1 + 3=4$.
• For $f(x)=-3\sin(x)$, when $x = 0$, $f(0)=-3\sin(0)=0$.
• For $f(x)=4\cos(x)$, when $x = 0$, $f(0)=4\cos(0)=4$.

Step2: Check minimum - value at $x=\frac{3\pi}{2}$ for remaining functions

• For $f(x)=\sin(x)+4$, when $x=\frac{3\pi}{2}$, $f(\frac{3\pi}{2})=\sin(\frac{3\pi}{2})+4=-1 + 4=3$.
• For $f(x)=\cos(x)+3$, when $x=\frac{3\pi}{2}$, $f(\frac{3\pi}{2})=\cos(\frac{3\pi}{2})+3=0 + 3=3$. But the derivative of $y = \cos(x)+3$ is $y'=-\sin(x)$, and $y'(\frac{3\pi}{2})=-\sin(\frac{3\pi}{2}) = 1>0$, so it is a local minimum.
• For $f(x)=4\cos(x)$, when $x=\frac{3\pi}{2}$, $f(\frac{3\pi}{2})=4\cos(\frac{3\pi}{2})=0$.

Answer:

$f(x)=\sin(x)+4$