What are the points of inflection, if any, of #f(x)= (pi/3)^((x^3-8) #?

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Answer 1 · 2017-01-08T04:02:34

#x =0 and x = -(2/(3ln(pi/3)))^(1/3)=-2.436#, nearly.

Taking logarithms and differentiating,

#(ln y)'=ln (pi/3)(x^3-8)'# that gives

#(y')/y=3ln(pi/3)x^2# and

#(y'')/y-((y')/y)^2=6xln(pi/3)# giving

#y''=0#, when# ((y')/y)^2=(3ln(pi/3)x^2)^2=-6x ln (pi/3)#

Solving, #x = 0 and x =-(2/(3ln(pi/3)))^(1/3)=-2.436#, nearly.,

graph{1.4461ln y-0.04612x^3=0 [-5, 5, 0, 0, 5]}.

Further differentiation for y''' reveals that

#y''' ne 0#, at these points.

So, the points of inflexion (POI) are

#x = 0 and x = -2.436#, nearly.

At these points, the tangent crosses the graph, to reverse its

sense of rotation, and change the form of the graph, from convexity

to concavity or vice versa.

See these effects in the inserted Socratic graph