find the points of inflection of the graph of the function. (if an answer does not exist, enter…

find the points of inflection of the graph of the function. (if an answer does not exist, enter dne.)\n$f(x)=x + 4\\cos(x),\\ 0,2\\pi$ \nsmaller $x$-value $(x,y)=(\\square)$ \nlarger $x$-value $(x,y)=(\\square)$ \ndetermine the open intervals on which the graph of the function is concave upward or concave downward. (enter your answers using interval notation. if an answer does not exist, enter dne.) \nconcave upward $\\square$ \nconcave downward $\\square$

find the points of inflection of the graph of the function. (if an answer does not exist, enter dne.)\n$f(x)=x + 4\\cos(x),\\ 0,2\\pi$ \nsmaller $x$-value $(x,y)=(\\square)$ \nlarger $x$-value $(x,y)=(\\square)$ \ndetermine the open intervals on which the graph of the function is concave upward or concave downward. (enter your answers using interval notation. if an answer does not exist, enter dne.) \nconcave upward $\\square$ \nconcave downward $\\square$

Answer

Explanation:

Step1: Find the first - derivative

Use the sum rule ((u + v)^\prime=u^\prime + v^\prime), where (u = x) and (v = 4\cos(x)). The derivative of (y=x) is (y^\prime=1), and the derivative of (y = 4\cos(x)) is (y^\prime=-4\sin(x)). So, (f^\prime(x)=1 - 4\sin(x)).

Step2: Find the second - derivative

Differentiate (f^\prime(x)) with respect to (x). The derivative of (1) is (0), and the derivative of (-4\sin(x)) is (-4\cos(x)). So, (f^{\prime\prime}(x)=-4\cos(x)).

Step3: Find the points of inflection

Set (f^{\prime\prime}(x) = 0), i.e., (-4\cos(x)=0). Then (\cos(x)=0). On the interval ([0,2\pi]), (x=\frac{\pi}{2}) and (x = \frac{3\pi}{2}). When (x=\frac{\pi}{2}), (y=\frac{\pi}{2}+4\cos(\frac{\pi}{2})=\frac{\pi}{2}). When (x=\frac{3\pi}{2}), (y=\frac{3\pi}{2}+4\cos(\frac{3\pi}{2})=\frac{3\pi}{2}).

Step4: Determine the concavity

  • For concave upward: (f^{\prime\prime}(x)>0), (-4\cos(x)>0), (\cos(x)<0). The solution on ([0,2\pi]) is ((\frac{\pi}{2},\frac{3\pi}{2})).
  • For concave downward: (f^{\prime\prime}(x)<0), (-4\cos(x)<0), (\cos(x)>0). The solution on ([0,2\pi]) is ([0,\frac{\pi}{2})\cup(\frac{3\pi}{2},2\pi]).

Answer:

smaller (x -)value ((x,y)=\left(\frac{\pi}{2},\frac{\pi}{2}\right)) larger (x -)value ((x,y)=\left(\frac{3\pi}{2},\frac{3\pi}{2}\right)) concave upward (\left(\frac{\pi}{2},\frac{3\pi}{2}\right)) concave downward (\left[0,\frac{\pi}{2}\right)\cup\left(\frac{3\pi}{2},2\pi\right])