Find the critical numbers of f(x)=2secx+tanx in 0

1 answer:

3 0

Using the range of [ 0 , 2π ]

$\bf f(x)=2sec(x)+tan(x)\implies \cfrac{df}{dx}=2sec(x)tan(x)+sec^2(x)
\\\\\\
0=2sec(x)tan(x)+sec^2(x)
\\\\\\
-sec^2(x)=2sec(x)tan(x)\implies \cfrac{-sec^2(x)}{2sec(x)}=tan(x)
\\\\\\
\cfrac{-sec(x)}{2}=tan(x)\implies \cfrac{-\frac{1}{cos(x)}}{\frac{2}{1}}=\cfrac{sin(x)}{cos(x)}
\\\\\\
-\cfrac{1}{2cos(x)}=\cfrac{sin(x)}{cos(x)}\implies -\cfrac{1}{2}=sin(x)\implies \measuredangle x =
\begin{cases}
\frac{7\pi }{6}\\\\
\frac{11\pi }{6}
\end{cases}$

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