1. ## Horizontal Tangent

1. f(x)= x+2cosx

2. f(x)= x+2sinx

3. f(x)= e^x (sinx)

Regarding problem number 1, I know that the derivative is 1-2sinx, then when I set it equal to zero I get sinx=1/2. Therefor it would be pi/6, but the answer key says that the answer is 2npi + pi/6, 2npi + 5pi/6. I need some clarification on how to reach that conclusion.

On problem 2, I get the answer cosx= -1/2. The answer key says that the final answer is (2n +1)pi +/- pi/3. Again I am quite sure how to reach that conclusion or final answer.

Same problem on 3. tanx= -1: What would my next step be to solve the problem.

Any feedback or help would be appreciated. These problems have been bugging me all day.

2. Originally Posted by Gitano
1. f(x)= x+2cosx

2. f(x)= x+2sinx

3. f(x)= e^x (sinx)

Regarding problem number 1, I know that the derivative is 1-2sinx, then when I set it equal to zero I get sinx=1/2. Therefor it would be pi/6, but the answer key says that the answer is 2npi + pi/6, 2npi + 5pi/6. I need some clarification on how to reach that conclusion.

This is just trigonometry: $\displaystyle \color{red}\sin x=\sin (\pi-x)\, \mbox{so}\,\sin x=\frac{1}{2}\Longrightarrow x=\frac{\pi}{6}+2n\pi$
$\displaystyle \color{red}\mbox{or}\,\, x=\pi-\frac{\pi}{6}+2n\pi=\frac{5\pi}{6}+2n\pi$

Remember that both basic trigonometry functions $\displaystyle \sin x\, \mbox{and} \cos x$ have period $\displaystyle =2\pi$

For the next two problems the issue is the same, only that $\displaystyle \tan x$ has period $\displaystyle \pi$

Tonio

On problem 2, I get the answer cosx= -1/2. The answer key says that the final answer is (2n +1)pi +/- pi/3. Again I am quite sure how to reach that conclusion or final answer.

Same problem on 3. tanx= -1: What would my next step be to solve the problem.

Any feedback or help would be appreciated. These problems have been bugging me all day.
..

3. Thanks for the clarification, appreciate it.

But I don't still don't quite understand problem 3. I get tan x = -1 and the answer key says that x= npi + pi/4. Tangent is negative in quadrants 2 and 4, so I don't understand how it is just pi/4.

4. Originally Posted by Gitano
Thanks for the clarification, appreciate it.

But I don't still don't quite understand problem 3. I get tan x = -1 and the answer key says that x= npi + pi/4. Tangent is negative in quadrants 2 and 4, so I don't understand how it is just pi/4.

I don't understand it either: $\displaystyle \tan x = -1 \Longrightarrow x=\frac{3\pi}{4}+n\pi$ , $\displaystyle n\in\mathbb{N}$

It must be a mistake, I suppose.

Tonio