Asked by Susana
Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line for y=(3^1/2)x+2cos(x), 0¡Üx<2¦Ð
Answers
Answered by
Reiny
For a horizontal tangent dy/dx = 0
dy/dx = √3 - 2sinx = 0
2sinx = √3
sinx = √3/2
x = 60° or 120°
x = π/3 or 2π/3
when x=π/3
y = √3(π/3) + 2cos(π/3) = √3(π/3) + 2(1/2) = (π√3 + 3)/3
when x = 2π/3
y = √3(2π/3) + 2(-1/2) = 2√3 π - 3)/3
dy/dx = √3 - 2sinx = 0
2sinx = √3
sinx = √3/2
x = 60° or 120°
x = π/3 or 2π/3
when x=π/3
y = √3(π/3) + 2cos(π/3) = √3(π/3) + 2(1/2) = (π√3 + 3)/3
when x = 2π/3
y = √3(2π/3) + 2(-1/2) = 2√3 π - 3)/3
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