Did you mean
y = x√(2-2cosx) or y = (√2)x - 2cosx ?
in either case, find dy/dx.
for a horizontal line, the slope, or dy/dx, is zero.
so set dy/dx = 0 and solve for x
Find those values of x at which the tangent line is horizontal to the curve
y=xsqrt2-2cosx
3 answers
The tangent to the graph of the y(x) function is horizontal when dy/dx = 0
I can't tell if your function is
y = x sqrt(2 - 2cosx) or
y = x*sqrt2 -2cosx
If it is the latter,
dy/dx = 0 when
sqrt2 = 2 sinx
sinx = (1/2)sqrt2
x = pi/4 and 3 pi/4
I can't tell if your function is
y = x sqrt(2 - 2cosx) or
y = x*sqrt2 -2cosx
If it is the latter,
dy/dx = 0 when
sqrt2 = 2 sinx
sinx = (1/2)sqrt2
x = pi/4 and 3 pi/4
(�ã2)x - 2cosx
1 answer