If #y = 1/(1+x^2)#, what are the points of inflection of the graph f (x)?

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Answer 1 · 2015-11-06T22:21:09

#x=pm 1/sqrt(3)#

The inflection points are the zeroes of the second derivative. So, first of all, we need to compute it:

#f(x)=1/(1+x^2)#

#f'(x)=-1/((1+x^2)^2) * 2x = - (2x)/((1+x^2)^2)#

#f''(x)= -\frac{2(1+x^2)^2-2x(2(1+x^2)*2x)}{(1+x^2)^4}#

#-\frac{2(1+x^2)^{cancel(2)} - 8x^2 cancel((1+x^2))}{(1+x^2)^{cancel(4) 3}#

#\frac{6x^2-2}{(1+x^2)^3}#

So, we need to find the zeroes of this function, which are the zeroes of its numerator:

#6x^2-2=0 iff 3x^2-1=0 iff x^2=1/3 iff x=pm 1/sqrt(3)#