+0  
 
0
53
6
avatar

Use the function and its derivative to determine any points on the graph of f at which the tangent line is horizontal. Use a graphing utility to verify your results. (If an answer does not exist, enter DNE.)

 

f(x) = 2 cos(x) − x,    f '(x) = −2 sin(x) − 1,    over the interval (0, 2𝜋)

 

  (smaller x-value)(x, y) = 

 

  (larger x-value)(x,y)=

 Mar 23, 2021
 #1
avatar+117787 
+2

-2sin x  -  1  = 0

 

-2sin x   =    1

 

sin x =  -1/2

 

This  is true at   7pi/6  and at 11pi/6

 

So

 

When x = 7pi/6, y=  2cos (7pi/6)  - pi/6  =   -sqrt (3)  - 7pi/6

 

When x = 11pi/6, y  = 2cos (11pi/6) - 11pi/6 =  sqrt (3) - 11pi/6

 

(x, y)    =  (7pi/6, -sqrt (3) - 7pi/6)   and  ( 11pi/6 , sqrt (3) - 11pi/6)

 

Here's a graph  :  https://www.desmos.com/calculator/eupogpqkpf

 

 

cool cool cool

 Mar 23, 2021
 #2
avatar+873 
0

What is a derivative?

 

=^._.^=

catmg  Mar 23, 2021
 #3
avatar+117787 
+3

In simple terms.....the first derivative is a function that  helps  you find the  slope of a  tangent line to a function at any point (if the function is differentiable at that point)....this is the  first derivative.....also used to find velocity

 

The second derivative is used to find mins and maxes of functions  (if they exist)  and also to calculate acceleration

 

I hope I've confused you enough, catmg.....LOL!!!!!

 

 

cool cool cool

CPhill  Mar 23, 2021
 #4
avatar+873 
0

I am very confused rn. 

Hopefully, I'll understand the concept in a few years. :)))

 

=^._.^=

catmg  Mar 23, 2021
 #5
avatar+117787 
+1

Don't worry.....you will learn this easily....your math skills won't give you any trouble in Calculus

 

 

cool cool cool

CPhill  Mar 23, 2021
 #6
avatar+873 
0

Awww, thank you. :))

 

=^._.^=

catmg  Mar 23, 2021

25 Online Users

avatar
avatar
avatar