Horizontal tangents to the graph of f(x) are tangent lines that have slope = 0, so to find these "critical points" for f, we take its derivative, set f'(x) = 0, and solve:
f(x) = x + 2 sinx
f'(x) = 1 + 2cosx = 0
cosx = - 1/2
x = 2π/3 or 4π/3
Van N.
asked 02/05/22Must only be the smallest positive value of x
Horizontal tangents to the graph of f(x) are tangent lines that have slope = 0, so to find these "critical points" for f, we take its derivative, set f'(x) = 0, and solve:
f(x) = x + 2 sinx
f'(x) = 1 + 2cosx = 0
cosx = - 1/2
x = 2π/3 or 4π/3
Find derivative of function. f''(x)=1+2cosx. Set this equal to zero and solve for x. cox=-1/2. General solution of x will be 2/3pi + (2pi)k and 4/3pi + (2pik), for all integer k.
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