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 the point (0,4)

Substitute (x = 0) into each function. For (y=\sin(x)+4), when (x = 0), (y=\sin(0)+4=0 + 4=4). For (y=\cos(x)+3), when (x = 0), (y=\cos(0)+3=1 + 3=4). For (y=-3\sin(x)), when (x = 0), (y=-3\sin(0)=0). For (y = 4\cos(x)), when (x = 0), (y=4\cos(0)=4). So we can't eliminate any function based on the point ((0,4)) yet.

Step2: Check the minimum - value point ((\frac{3\pi}{2},3))

For (y=\sin(x)+4), when (x=\frac{3\pi}{2}), (y=\sin(\frac{3\pi}{2})+4=-1 + 4=3). For (y=\cos(x)+3), when (x=\frac{3\pi}{2}), (y=\cos(\frac{3\pi}{2})+3=0 + 3=3). For (y=-3\sin(x)), when (x=\frac{3\pi}{2}), (y=-3\sin(\frac{3\pi}{2})=-3\times(-1)=3). For (y = 4\cos(x)), when (x=\frac{3\pi}{2}), (y=4\cos(\frac{3\pi}{2})=0). So we can eliminate (y = 4\cos(x)).

Step3: Analyze the nature of the functions at the point

The derivative of (y=\sin(x)+4) is (y'=\cos(x)). At (x = \frac{3\pi}{2}), (y'=\cos(\frac{3\pi}{2})=0), and the second - derivative (y''=-\sin(x)), at (x=\frac{3\pi}{2}), (y''=-\sin(\frac{3\pi}{2}) = 1>0), so it has a minimum at (x=\frac{3\pi}{2}). The derivative of (y=\cos(x)+3) is (y'=-\sin(x)). At (x=\frac{3\pi}{2}), (y'=-\sin(\frac{3\pi}{2}) = 1\neq0), so it does not have a minimum at (x=\frac{3\pi}{2}). The derivative of (y=-3\sin(x)) is (y'=-3\cos(x)). At (x=\frac{3\pi}{2}), (y'=-3\cos(\frac{3\pi}{2})=0), and the second - derivative (y'' = 3\sin(x)), at (x=\frac{3\pi}{2}), (y''=3\sin(\frac{3\pi}{2})=-3<0), so it has a maximum at (x=\frac{3\pi}{2}).

Answer:

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