Respuesta :
∴ F is concave up on [-2π/3, 0) , F is concave down on (0,2π/3]
Inflection points are points on a graph where the curvature of the g[tex]f(x)= y = x+sin2x\\f'(x) = 1+cos2x\\f"(x) = -4sin2x\\f"(x) = 0= -4sin2x=0\\sin2x = 0\\2x= sin^{-1}(0)\\2x = 0\\x= 0\\[/tex]
There is an inflection point at x=0
[tex]f'(x)=0 = 1+2cos2x=0\\2cos2x =-1\\cos2x = \frac{-1}{2} \\2x = cos^{-1} (\frac{-1}{2})\\[/tex]
2x = 2π/3
x = π/3
There is only one critical point, x= π/3
f"(π/3) = -4 × sin(2π/3)
= -4 × [tex]\frac{\sqrt{3} }{2}[/tex]
= [tex]-2\sqrt{3}[/tex] <0
There is one local maximum at x= π/3.
There is no local minimum.
There is an inflection point at x = 0 is 0.
F"(-π/4) =4sin (2x -π/4)
= -4 sin (-π/2)
= 4sin(π/2)
= 4>0
∴ F is concave up on [-2π/3, 0)
f" (π/4) = -4sin (2 × π/4)
= -4<0
∴ F is concave down on (0,2π/3]
Inflection points are points on a graph where the curvature of the graph changes from concave to convex or vice versa. For example, in a quadratic equation, the inflection point is where the graph changes from having a 'U' shape to having an 'inverse U' shape. In other words, the point at which the graph changes from being concave down to being concave up (or vice versa). Inflection points can also occur in higher-order polynomials and in trigonometric functions.
In addition, inflection points can be used to identify local extrema or points at which the function attains either a minimum or maximum value. In general, an inflection point is a point at which the concavity of a function changes. This can be observed by examining the second derivative of the function at the point: if the second derivative is positive, then the function has a convex curvature; if the second derivative is negative, then the function has a concave curvature.
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