Answer to Question #273991 in Calculus for Josa Kyme Palma

Question #273991

Determine the open intervals on which the graph of f(x)= (x²)/(x²+1) is concave upward or concave downward


1
Expert's answer
2021-12-02T16:57:38-0500
f(x)=x2x2+1f(x)=\dfrac{x^2}{x^2+1}

Domain: (,)(-\infin, \infin)


f(x)=2x(x2+1)2x(x2)(x2+1)2=2x(x2+1)2f'(x)=\dfrac{2x(x^2+1)-2x(x^2)}{(x^2+1)^2}=\dfrac{2x}{(x^2+1)^2}

f(x)=2((x2+1)22x(2x)(x2+1))(x2+1)4f''(x)=\dfrac{2((x^2+1)^2-2x(2x)(x^2+1))}{(x^2+1)^4}

=2(x2+14x2)(x2+1)3=2(13x2)(x2+1)3=\dfrac{2(x^2+1-4x^2)}{(x^2+1)^3}=\dfrac{2(1-3x^2)}{(x^2+1)^3}

Find the point(s) of inflection


f(x)=0=>2(13x2)(x2+1)3=0f''(x)=0=>\dfrac{2(1-3x^2)}{(x^2+1)^3}=0

13x2=01-3x^2=0

x1=33,x2=33x_1=-\dfrac{\sqrt{3}}{3}, x_2=\dfrac{\sqrt{3}}{3}

If x<33,f(x)<0,f(x)x<-\dfrac{\sqrt{3}}{3}, f''(x)<0, f(x) is concave downward.

If 33<x<33,f(x)>0,f(x)-\dfrac{\sqrt{3}}{3}<x<\dfrac{\sqrt{3}}{3}, f''(x)>0, f(x) is concave upward.

If x>33,f(x)<0,f(x)x>\dfrac{\sqrt{3}}{3}, f''(x)<0, f(x) is concave downward.


f(x)f(x) is concave upward on (33,33).(-\dfrac{\sqrt{3}}{3}, \dfrac{\sqrt{3}}{3}).

f(x)f(x) is concave downward on (,33)(33,).(-\infin,-\dfrac{\sqrt{3}}{3})\cup ( \dfrac{\sqrt{3}}{3}, \infin).



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