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

#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}#

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

To find the points of inflection of the graph of \( f(x) = \frac{1}{1+x^2} \), we need to find where the concavity changes. 1. Find the second derivative of \( f(x) \): \[ f''(x) = \frac{d^2}{dx^2} \left(\frac{1}{1+x^2}\right) \] \[ f''(x) = \frac{d}{dx} \left( \frac{-2x}{(1+x^2)^2} \right) \] \[ f''(x) = \frac{-2(1+x^2)^2 - 2x(2)(1+x^2)(2x)}{(1+x^2)^4} \] \[ f''(x) = \frac{-2(1+x^2)(1+x^2 - 4x^2)}{(1+x^2)^4} \] \[ f''(x) = \frac{-2(1+x^2-3x^2)}{(1+x^2)^3} \] \[ f''(x) = \frac{-2(1-2x^2)}{(1+x^2)^3} \] 2. To find the points of inflection, set \( f''(x) = 0 \) and solve for \( x \): \[ -2(1-2x^2) = 0 \] \[ 1 - 2x^2 = 0 \] \[ 2x^2 = 1 \] \[ x^2 = \frac{1}{2} \] \[ x = \pm \frac{1}{\sqrt{2}} \] 3. Check the concavity on intervals separated by these points to determine if they are points of inflection. - For \( x < -\frac{1}{\sqrt{2}} \), \( f''(x) > 0 \) (concave up) - For \( -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \), \( f''(x) < 0 \) (concave down) - For \( x > \frac{1}{\sqrt{2}} \), \( f''(x) > 0 \) (concave up) Therefore, the points of inflection are \( x = -\frac{1}{\sqrt{2}} \) and \( x = \frac{1}{\sqrt{2}} \).