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

Question

consider the following function.

$y = 7x - 3\tan(x)$, $\left(-\frac{\pi}{2},\frac{\pi}{2}\
ight)$

find the first and second derivatives.

$y(x) =$

$y(x) =$

find 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.)

$c =$

determine 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.)

concave upward

concave downward

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 \(\sec^{2}x=\frac{1}{\cos^{2}x}
eq0\) 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: \((0,\frac{\pi}{2})\)