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

the graph of which function passes through (0,4) and has a minimum value at $\\left(\\frac{3\\pi}{2},3\\right)$?\n$\\bigcirc f(x)=\\sin(x)+4$\n$\\bigcirc f(x)=\\cos(x)+3$\n$\\bigcirc f(x)= - 3\\sin(x)$\n$\\bigcirc f(x)=4\\cos(x)$

the graph of which function passes through (0,4) and has a minimum value at $\\left(\\frac{3\\pi}{2},3\\right)$?\n$\\bigcirc f(x)=\\sin(x)+4$\n$\\bigcirc f(x)=\\cos(x)+3$\n$\\bigcirc f(x)= - 3\\sin(x)$\n$\\bigcirc f(x)=4\\cos(x)$

Answer

Explanation:

Step1: Check the 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). So (f(x)=-3\sin(x)) is eliminated as it does not pass through ((0,4)).

Step2: Check the minimum - value point ((\frac{3\pi}{2},3)) for the 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).
  • For (f(x)=4\cos(x)), when (x=\frac{3\pi}{2}), (f(\frac{3\pi}{2})=4\cos(\frac{3\pi}{2})=0). So (f(x)=4\cos(x)) is eliminated.

Step3: Analyze the behavior of the remaining functions

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

Answer:

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