consider the equation below. (if an answer does not exist, enter dne.)\n$f(x)=5\\sin (x)+5\\cos (x),\\…

consider the equation below. (if an answer does not exist, enter dne.)\n$f(x)=5\\sin (x)+5\\cos (x),\\ 0\\leq x\\leq 2\\pi$\n(a) find the interval on which $f$ is increasing. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(b) find the local minimum and maximum values of $f$.\nlocal minimum value\nlocal maximum value\n(c) find the inflection points.\n$(x,y)=$(smaller $x$-value)\n$(x,y)=$(larger $x$-value)\nfind the interval on which $f$ is concave up. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is concave down. (enter your answer using interval notation.)\n

consider the equation below. (if an answer does not exist, enter dne.)\n$f(x)=5\\sin (x)+5\\cos (x),\\ 0\\leq x\\leq 2\\pi$\n(a) find the interval on which $f$ is increasing. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is decreasing. (enter your answer using interval notation.)\n\n(b) find the local minimum and maximum values of $f$.\nlocal minimum value\nlocal maximum value\n(c) find the inflection points.\n$(x,y)=$(smaller $x$-value)\n$(x,y)=$(larger $x$-value)\nfind the interval on which $f$ is concave up. (enter your answer using interval notation.)\n\nfind the interval on which $f$ is concave down. (enter your answer using interval notation.)\n

Answer

Explanation:

Step1: Find the first derivative

The derivative of (f(x)=5\sin(x)+5\cos(x)) is (f^\prime(x) = 5\cos(x)-5\sin(x)). Set (f^\prime(x)=0), then (5\cos(x)-5\sin(x)=0), which simplifies to (\tan(x) = 1). The solutions in (0\leq x\leq2\pi) are (x=\frac{\pi}{4}) and (x=\frac{5\pi}{4}).

Step2: Determine increasing and decreasing intervals

Test intervals:

  • For (0\leq x<\frac{\pi}{4}), let (x = 0), (f^\prime(0)=5>0), so (f(x)) is increasing on ([0,\frac{\pi}{4})).
  • For (\frac{\pi}{4}<x<\frac{5\pi}{4}), let (x=\frac{\pi}{2}), (f^\prime(\frac{\pi}{2})=- 5<0), so (f(x)) is decreasing on ((\frac{\pi}{4},\frac{5\pi}{4})).
  • For (\frac{5\pi}{4}<x\leq2\pi), let (x=\frac{3\pi}{2}), (f^\prime(\frac{3\pi}{2})=5>0), so (f(x)) is increasing on ((\frac{5\pi}{4},2\pi]).

Step3: Find local extrema

  • Local maximum: (f(\frac{\pi}{4})=5\sqrt{2}).
  • Local minimum: (f(\frac{5\pi}{4})=-5\sqrt{2}).

Step4: Find the second derivative

(f^{\prime\prime}(x)=-5\sin(x)-5\cos(x)). Set (f^{\prime\prime}(x) = 0), then (\sin(x)+\cos(x)=0), (\tan(x)=-1). The solutions in (0\leq x\leq2\pi) are (x=\frac{3\pi}{4}) and (x=\frac{7\pi}{4}). (f(\frac{3\pi}{4}) = 0), (f(\frac{7\pi}{4})=0).

Step5: Determine concavity

  • Test intervals:
    • For (0\leq x<\frac{3\pi}{4}), let (x = 0), (f^{\prime\prime}(0)=-5<0), concave down on ([0,\frac{3\pi}{4})).
    • For (\frac{3\pi}{4}<x<\frac{7\pi}{4}), let (x=\pi), (f^{\prime\prime}(\pi)=5>0), concave up on ((\frac{3\pi}{4},\frac{7\pi}{4})).
    • For (\frac{7\pi}{4}<x\leq2\pi), let (x = 2\pi), (f^{\prime\prime}(2\pi)=-5<0), concave down on ((\frac{7\pi}{4},2\pi]).

Answer:

(a) Increasing: ([0,\frac{\pi}{4})\cup(\frac{5\pi}{4},2\pi]); Decreasing: ((\frac{\pi}{4},\frac{5\pi}{4})) (b) Local minimum value: (-5\sqrt{2}); Local maximum value: (5\sqrt{2}) (c) Inflection points: ((\frac{3\pi}{4},0)) (smaller (x)-value), ((\frac{7\pi}{4},0)) (larger (x)-value); Concave up: ((\frac{3\pi}{4},\frac{7\pi}{4})); Concave down: ([0,\frac{3\pi}{4})\cup(\frac{7\pi}{4},2\pi])