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)

For a function (y = f(x)), substitute (x = 0) into each option.

• Option 1: (f(x)=\sin(x)+4), when (x = 0), (f(0)=\sin(0)+4=0 + 4=4).
• Option 2: (f(x)=\cos(x)+3), when (x = 0), (f(0)=\cos(0)+3=1 + 3=4).
• Option 3: (f(x)=-3\sin(x)), when (x = 0), (f(0)=-3\sin(0)=0).
• Option 4: (f(x)=4\cos(x)), when (x = 0), (f(0)=4\cos(0)=4\times1 = 4). So we can eliminate the third - option.

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

• Option 1: (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).
• Option 2: (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).
• Option 4: (f(x)=4\cos(x)), when (x=\frac{3\pi}{2}), (f(\frac{3\pi}{2})=4\cos(\frac{3\pi}{2})=0). We know that the derivative of (y = \sin(x)+4) is (y'=\cos(x)), and (\cos(\frac{3\pi}{2}) = 0). The second - derivative (y''=-\sin(x)), and (y''(\frac{3\pi}{2})=-\sin(\frac{3\pi}{2}) = 1>0), so (y=\sin(x)+4) has a minimum at (x = \frac{3\pi}{2}). The derivative of (y=\cos(x)+3) is (y'=-\sin(x)), and (-\sin(\frac{3\pi}{2})=1\neq0), so (y = \cos(x)+3) does not have a minimum at (x=\frac{3\pi}{2}).

Answer:

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