consider the following function.\n\n$y = 7x - 3\\tan(x)$, $\\left(-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right)$\n…

consider the following function.\n\n$y = 7x - 3\\tan(x)$, $\\left(-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right)$\n\nfind the first and second derivatives.\n\n$y(x) =$\n\n$y(x) =$\n\nfind any values of $c$ such that $y(c) = 0$. (enter your answer as a comma - separated list. if any answer does not exist, enter dne.)\n\n$c =$\n\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.)\n\nconcave upward\n\nconcave downward

consider the following function.\n\n$y = 7x - 3\\tan(x)$, $\\left(-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right)$\n\nfind the first and second derivatives.\n\n$y(x) =$\n\n$y(x) =$\n\nfind any values of $c$ such that $y(c) = 0$. (enter your answer as a comma - separated list. if any answer does not exist, enter dne.)\n\n$c =$\n\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.)\n\nconcave upward\n\nconcave downward

Answer

Explanation:

Step1: Find the first derivative

Use the sum rule ((u - v)^\prime=u^\prime - v^\prime) and the basic derivative rules ((x^n)^\prime=nx^{n - 1}) and ((\tan x)^\prime=\sec^{2}x). For (y = 7x-3\tan(x)), (y^\prime(x)=(7x)^\prime-(3\tan x)^\prime). Since ((7x)^\prime = 7) and ((3\tan x)^\prime=3\sec^{2}x), then (y^\prime(x)=7 - 3\sec^{2}x).

Step2: Find the second derivative

Differentiate (y^\prime(x)=7 - 3\sec^{2}x) with respect to (x). Use the chain - rule ((u^{n})^\prime=nu^{n - 1}u^\prime), where (u = \sec x), (n = 2). ((\sec x)^\prime=\sec x\tan x). (y^{\prime\prime}(x)=-3\times2\sec x\times(\sec x\tan x)=-6\sec^{2}x\tan x).

Step3: Solve (y^{\prime\prime}(c) = 0)

Set (y^{\prime\prime}(x)=-6\sec^{2}x\tan x = 0). Since (\sec^{2}x=\frac{1}{\cos^{2}x}\neq0) for (x\in(-\frac{\pi}{2},\frac{\pi}{2})), then (\tan x = 0). (\tan x=\frac{\sin x}{\cos x}=0) when (\sin x = 0) and (x\in(-\frac{\pi}{2},\frac{\pi}{2})), so (x = 0). Thus (c = 0).

Step4: Determine concavity

Test intervals ((-\frac{\pi}{2},0)) and ((0,\frac{\pi}{2})). Take a test point (x=-\frac{\pi}{4}) in ((-\frac{\pi}{2},0)): (y^{\prime\prime}(-\frac{\pi}{4})=-6\sec^{2}(-\frac{\pi}{4})\tan(-\frac{\pi}{4})=-6\times2\times(-1)=12>0). Take a test point (x=\frac{\pi}{4}) in ((0,\frac{\pi}{2})): (y^{\prime\prime}(\frac{\pi}{4})=-6\sec^{2}(\frac{\pi}{4})\tan(\frac{\pi}{4})=-6\times2\times1=-12<0).

Answer:

(y^\prime(x)=7 - 3\sec^{2}x) (y^{\prime\prime}(x)=-6\sec^{2}x\tan x) (c = 0) concave upward: ((-\frac{\pi}{2},0)) concave downward: ((0,\frac{\pi}{2}))