I have a calculus question...?

Question

How would I find the x coordinates of all the points on the curve y=sin2x-2sinx at which the tangent line is horizontal?

Answer 1

The derivative of the function given is

y'=cos(2x)*2 - 2*cos(x) = 2(2cos^2(x)-1)-2*cos(x)
= 4cos^2(x)-2cos(x)-2 = 2(2cos^2(x)-cos(x)-1)
=2(2cos(x)+1)(cos(x)-1)

For y'=0, either 2cos(x)+1=0 or cos(x)-1=0, from which we get

cos(x)=-1/2 (which gives either x=2*pi/3 + 2*pi*k, k an integer or x=-2*pi/3 + 2*pi*k, k an integer)

or
cos(x)=1, so x=2*pi*k, k an integer.

Answer 2

x cordinate is 0 means that the given line is a tangent..or y=0 for those points..also the slope is 0..
so differentiate the given function..set it to 0 and solve for x values..

Answer 3

i'd say structure is component to engineering which signifies that a good draw close on math is major, pre-cal is is like algebra to the subsequent factor with the Pi chart and understanding about radians and perspective measures (which seems major) yet Calculus is the learn of derivatives, (the slope of a line at a particular factor) type of unnecessary whatever field of workd your going to, in case you inquire from me.

Answer 4

f(x)=sin(2x)-2sin(x)
f'(x)=2cos(2x)-2cos(x)

Set f'(x)=0 and solve for x. This gives you -2.094, 0, and 2.094.

Now plug those values back into the original equation, f(x), and that gives you your horizontal lines with a slope of 0.

f(-2.094)=2.59808
f(0)=0
f(2.094)=-2.59808

If you graph f(x), you see the horizontal tangent lines are indeed at the values above.

Visit the source below to see the graph:

Answer 5

First, you must differentiate this function for the first derivative.

f'(x)=2cos(2x)

Set this equal to 0 and solve for x. The solution is Pi/4 or approximately 3.14/4