What is the x-coordinate of the point of inflection on the graph of y=1/10x^(5)+1/2X^(4)-3/10?

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We meet the Curvature Points of ##y## by meeting the promote derivative of the capacity (y''), and the x-values at which y'' equals 0.

We behold for the zeroes consequently at those points the angularity (or the line in which the mount of the capacity ##f(x)## is trending) has leveled off; it is at these points that the angularity is most mitigated to depend from dogmatic to indirect, or vice-versa.

Just as enhancement, Math is Fun offers notes on curvature points.

##y= 1/10x^5 +1/2x^4 -3/10## ##y'=1/2x^4+ 2x^3## ##y''=2x^3+6x^2##

We set our promote derivative to 0: ##y''=x^2(2x+6)=0##

And we can meet our curvature points through these equations:

Our Curvature Points are then at ##x= 0, and x=-3##

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