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)=
+128448 -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
+128448 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!!!!!
+2401 I am very confused rn.
Hopefully, I'll understand the concept in a few years. :)))
=^._.^=
