The Graph of f
Date: Wed, 3 Jul 1996 18:22:11 -0400 (EDT) From: Anonymous Subject: function f(x)=3x^(1/3)-x Consider the function f(x)=3x^(1/3)-x. Where are its local and absolute maxima and minima? What are the x-intercepts of the graph of f? Does its graph have inflection points? Are there asymptotes? Where is the graph concave upward or downward? Where is the function increasing/ decreasing? Does the graph exhibit any symmetries? Sketch the graph. Thanking you ahead of time, Dr. Math.
Date: Thu, 4 Jul 1996 12:43:49 -0400 (EDT) From: Dr. Anthony Subject: Re: function f(x)=3x^(1/3)-x We note that this is an odd function of x, and so will have point symmetry about the origin. f'(x) = 3(1/3)x^(-2/3) - 1 = 1/x^(2/3) - 1 So f'(x) = 0 when x^(2/3) = 1, that is x = 1 or x = -1 The f(1) = 3-1 = 2 and f(-1) = -3 -(-1) = -2 So turning points at (1,2) (max.point) and (-1,-2) (min.point) When x=0 then y=0, so curve passes through the origin, and f'(x) is infinite, so curve is vertical at the origin When y=0 we also have 3x^(1/3) = x so 3 = x^(2/3) or x = 3^(3/2) x = +or- 5.196 The curve cuts the x axis at +and- 5.196 When x -> infinity f'(x) -> 1/infin - 1 -> -1, so the curve tends to a slope -1 as x increases. The curve has an S shape between (-1,-2), (0,0) and (1,2), the first and last of these being turning points, and the origin (0,0), a point where the curve is vertical. From (1,2) the curve slopes down as x increases, crosses the x axis at 5.196 and tends to a slope -1 as x tends to infinity . From (-1,-2) the curve bends up as x decreases, crosses the x axis at -5.196 and tends to a slope -1 as x tends to -infinity. There is one point of inflection (f''(x)=0) and that is at the origin. There are no asymptotes. -Doctor Anthony, The Math Forum
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