Determine the points on the graph of y= cos2x - cos x,x E [0,2pi], where the tangent is horizontal .
Determine the equation of the tangent line to the graph of y= f(x)= x(2^-x) at (0,0).
y' = -2sin2x + sinx
so, where is y'=0?
2sin2x - sinx = 0
2sinx cosx - sinx = 0
sinx(2cosx - 1) = 0
sinx = 0
cosx = 1/2
Now supply the 4 angles (actually, 5, since you have a closed interval)
y = x 2^-x
using the product and chain rule,
y' = 2^-x + x * ln2 2^-x (-1) = 2^-x (1 - x ln2)
y'(0) = 1(1-0) = 1
so the line is y=x
confirm at
https://www.wolframalpha.com/input/?i=plot+y%3Dx+2%5E-x%2C+y%3Dx
To find the points on the graph of the equation y = cos(2x) - cos(x) where the tangent is horizontal, we need to find the critical points of the graph.
A horizontal tangent occurs when the derivative of the function is equal to zero. Hence, we need to find the derivative of the function y = cos(2x) - cos(x).
Step 1: Find the derivative of y with respect to x.
By using the chain rule, the derivative of cos(2x) is -2sin(2x) and the derivative of cos(x) is -sin(x).
Therefore, the derivative of y is dy/dx = -2sin(2x) + sin(x).
Step 2: Set the derivative equal to zero and solve for x.
-2sin(2x) + sin(x) = 0.
To solve this equation, we can factor out sin(x):
sin(x)(1 - 2cos(x)) = 0.
This equation will be satisfied if either sin(x) = 0 or 1 - 2cos(x) = 0.
For sin(x) = 0, the values of x that satisfy this condition are x = 0, π, and 2π.
For 1 - 2cos(x) = 0, we need to solve for cos(x):
2cos(x) = 1.
cos(x) = 1/2.
The solutions for cos(x) = 1/2 occur at x = π/3 and x = 5π/3.
Therefore, the points on the graph where the tangent is horizontal correspond to the values of x = 0, π, 2π, π/3, and 5π/3.
To find the corresponding y-values, substitute these x-values back into the original equation y = cos(2x) - cos(x):
For x = 0: y = cos(2*0) - cos(0) = 1 - 1 = 0.
For x = π: y = cos(2*π) - cos(π) = 1 - (-1) = 2.
For x = 2π: y = cos(2*2π) - cos(2π) = 1 - 1 = 0.
For x = π/3: y = cos(2*π/3) - cos(π/3) = -1/2 - √3/2.
For x = 5π/3: y = cos(2*5π/3) - cos(5π/3) = -1/2 - √3/2.
Therefore, the points on the graph where the tangent is horizontal are (0, 0), (π, 2), (2π, 0), (π/3, -1/2 - √3/2), and (5π/3, -1/2 - √3/2).