Chapter 3.5, Problem 27E

To determine

To show:That the graph of $y=\sec x$ and $y=co\mathrm{s}x$ have horizontal tangents at $x=0$ .

Answer to Problem 27E

Both equations have horizontal tangents at $x=0$ becausederivative is $0$ at $x=0$ .

Explanation of Solution

Given information:Given equation is $y=\sec x$ and $y=co\mathrm{s}x$ .

Calculation:Given equation is $y=\sec x$ and $y=co\mathrm{s}x$ .

  $\begin{array}{l}y=\sec x\\ \frac{dy}{dx}=\sec x\tan x\\ \mathrm{put}x=0\mathrm{in}\frac{dy}{dx}\\ \frac{dy}{dx}=\sec 0.\tan 0=1\times 0=0\end{array}$

  $y=\sec x$ has horizontal tangent at $x=0$ because derivative is $0$ at $x=0$ .

  $\begin{array}{l}y=co\mathrm{s}x\\ \frac{dy}{dx}=-\sin x\\ \mathrm{put}x=0\mathrm{in}\frac{dy}{dx}\\ \frac{dy}{dx}=-\sin 0=0\end{array}$

  $y=co\mathrm{s}x$ has horizontal tangent at $x=0$ because derivative is $0$ at $x=0$ .