Determine the points on the graph of y= cos2x - cos x,x E [0,2pi], where the tangent is horizontal .
3 answers
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)
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
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