Monday, April 27, 2015

Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 33

Find what values of $x$ does the graph of $f(x) = x + 2 \sin x$ have a horizontal tangent.

Solving for $f'(x)$


$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{d}{dx} (x) + 2 \frac{d}{dx} (\sin x)
&& \text{}
\\
\\
f'(x) =& 1 + 2 \cos x
&& \text{}
\\
\\
m_T =& 0 \qquad \text{ slope of the tangent is horizontal}
&&
\\
\\


\end{aligned}
\end{equation}
$


Let $f'(x) = m_T$ (slope of the tangent line)


$
\begin{equation}
\begin{aligned}

f'(x) = m_T =& 1 + 2 \cos x
\\
\\
0 =& 1 + 2 \cos x
\\
\\
2 \cos x =& -1
\\
\\
2 \cos x =& \frac{-1}{2}


\end{aligned}
\end{equation}
$


By using the unit circle diagram, we can determine what angle(s) has $\displaystyle \frac{-1}{2}$ on $x$-coordinate, so..


$
\begin{equation}
\begin{aligned}

x =& \cos^{-1} \left[ \frac{-1}{2} \right]
\\
\\
x =& \frac{2}{3} \pi \text{ and } x = \frac{4}{3} \pi

\end{aligned}
\end{equation}
$


Also, we know that the trigonometric functions have repeating cycles so the answer is

$\displaystyle x = \frac{4}{3} \pi + 2 \pi (n)$ and $\displaystyle x = \frac{2}{3} \pi + 2 \pi (n) $ ; where $n$ is any integer and $2 \pi$ corresponds to the repeating period.

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