Find those values of x at which the tangent line is horizontal to the curve
y=xsqrt2-2cosx
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
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
(�ã2)x - 2cosx
To find the values of x at which the tangent line is horizontal to the curve y = xsqrt(2) - 2cos(x), we need to find the derivative of the curve and set it equal to zero.
Step 1: Find the derivative of the curve y = xsqrt(2) - 2cos(x) with respect to x.
To find the derivative, differentiate each term separately. The derivative of xsqrt(2) is sqrt(2), and the derivative of -2cos(x) is 2sin(x). Therefore, the derivative of y is:
dy/dx = sqrt(2) - 2sin(x)
Step 2: Set the derivative equal to zero and solve for x.
To find the values of x at which the tangent line is horizontal, we set the derivative equal to zero and solve the resulting equation:
sqrt(2) - 2sin(x) = 0
First, isolate the sin(x) term by moving sqrt(2) to the other side:
2sin(x) = sqrt(2)
Divide both sides of the equation by 2:
sin(x) = sqrt(2) / 2
Step 3: Determine the values of x that satisfy the equation.
To find the values of x that satisfy sin(x) = sqrt(2) / 2, we can look at the unit circle or use inverse trigonometric functions.
The solutions for sin(x) = sqrt(2) / 2 are x = π/4 + 2πn and x = 3π/4 + 2πn for any integer value n.
So, the values of x at which the tangent line is horizontal to the curve y = xsqrt(2) - 2cos(x) are x = π/4 + 2πn and x = 3π/4 + 2πn, where n is an integer.