+209 (a) Find all points on the graph of ff where the tangent line is horizontal.
(b) Graph the function and the horizontal tangent lines on the interval [−2π,2π][−2π,2π] on the same screen.
f(x)=2sinx+cosx
I honestly have no idea how to do this..the answer looks crazy too. Any help is appreciated.
+129852 f(x)=2sinx+cosx
1st....find the derivative
f ' (x) = 2cosx - sinx
Set this = 0........the solutions will tell us where the derivative = 0, that is, where the tangent lines are horizontal
2cosx - sinx = 0 square both sides.......this may produce some "extraneous" solutions......we'll deal wth this possibility in a sec
4cos^2x = sin^2x and notce that we can write
4cos^2x = 1 - cos^2x add cos^2x to both sides
5cos^2x = 1 divide both sides by 5
cos^2x = 1/5 take the square root of both sides
cosx = +1/ √5 or cos x = -1 / √5
Let's deal with the first possibility......I'm going to work with degrees....we'll covert back to radians
cosx = +1/ √5 and this happens at about 63.4° and at about 296.56°
And the second solution is
cosx = -1/ √5 and this happens at about about 116.56° and at about 243.43°
Now......look at the graph [in degrees] : https://www.desmos.com/calculator/dokwgbq8vg
Notice that the only two places from [0 to 360] that the tangent lines are horizontal occur at about 63.4° and at about 243.43°
So the other two solutions are extraneous [ produced because we squared the original equation]
And the general solutions on [0, 360] degrees are
x = 63.4° + n 360° and x = 243.43° + n 360° where n is an integer
So.....converting to radians we have
x = 1.1065 + n(2pi) and x = 4.248 +n (2pi) where n is an integer
So......the correct answers in radian measure on [0, 2pi] are
x = 1.1065 rads and x = 4.248 rads on [ 0, 2p]
And ....since the answer also specified the interval [-2pi, 0], we have
x = 1.1065 - 2pi = -5.177rads and x = 4.248 - 2pi = -2.034 rads on [-2pi, 0]
Here's the graph in rads : https://www.desmos.com/calculator/apqa44zvad
