Determine the open intervals on which the graph of the function
$f ( x ) = \frac { x ^ { 2 } } { x ^ { 2 } + 625 }$is concave upward or concave downward.
A) concave upward: $\left( - \infty , - \frac { 25 \sqrt { 3 } } { 3 } \right) , \left( \frac { 25 \sqrt { 3 } } { 3 } , \infty \right) ;$ concave downward: $\left( - \frac { 25 \sqrt { 3 } } { 3 } , \frac { 25 \sqrt { 3 } } { 3 } \right)$
B) concave upward: $\left( - \frac { 25 \sqrt { 3 } } { 3 } , \frac { 25 \sqrt { 3 } } { 3 } \right) ;$ concave downward: $\left( - \infty , - \frac { 25 \sqrt { 3 } } { 3 } \right) , \left( \frac { 25 \sqrt { 3 } } { 3 } , \infty \right)$
C) concave upward: $( - \infty , 0 )$; concave downward: $( 0 , \infty )$
D) concave upward: $\left( - \frac { 625 } { 3 } , \frac { 625 } { 3 } \right) ;$ concave downward: $\left( - \infty , - \frac { 625 } { 3 } \right) , \left( \frac { 625 } { 3 } , \infty \right)$
E) concave upward: $\left( - \infty , - \frac { 625 } { 3 } \right) , \left( \frac { 625 } { 3 } , \infty \right) ;$ concave downward: $\left( - \frac { 625 } { 3 } , \frac { 625 } { 3 } \right)$

Question

Determine the open intervals on which the graph of the function 
$f ( x ) = \frac { x ^ { 2 } } { x ^ { 2 } + 625 }$is concave upward or concave downward. 
A) concave upward: $\left( - \infty , - \frac { 25 \sqrt { 3 } } { 3 } \right) , \left( \frac { 25 \sqrt { 3 } } { 3 } , \infty \right) ;$ concave downward: $\left( - \frac { 25 \sqrt { 3 } } { 3 } , \frac { 25 \sqrt { 3 } } { 3 } \right)$
B) concave upward: $\left( - \frac { 25 \sqrt { 3 } } { 3 } , \frac { 25 \sqrt { 3 } } { 3 } \right) ;$ concave downward: $\left( - \infty , - \frac { 25 \sqrt { 3 } } { 3 } \right) , \left( \frac { 25 \sqrt { 3 } } { 3 } , \infty \right)$
C) concave upward: $( - \infty , 0 )$; concave downward: $( 0 , \infty )$
D) concave upward: $\left( - \frac { 625 } { 3 } , \frac { 625 } { 3 } \right) ;$ concave downward: $\left( - \infty , - \frac { 625 } { 3 } \right) , \left( \frac { 625 } { 3 } , \infty \right)$
E) concave upward: $\left( - \infty , - \frac { 625 } { 3 } \right) , \left( \frac { 625 } { 3 } , \infty \right) ;$ concave downward: $\left( - \frac { 625 } { 3 } , \frac { 625 } { 3 } \right)$

Quizplus · Accepted Answer

To determine the intervals of concavity, we need to find the second derivative of the function: $$f(x)=\frac{x^2}{x^2+625}$$ $$f'(x)=\frac{625}{(x^2+625)^2}(2x(x^2+625)-x^2(2x))=\frac{625x^2}{(x^2+625)^2}$$ $$f''(x)=\frac{625(3x^4-2500x^2+3125)}{(x^2+625)^3}$$ To find the intervals of concavity, we need to solve $f''(x)=0$, or $3x^4-2500x^2+3125=0$. This can be factored as $125(5x^2-25\sqrt{3}x+125)(5x^2+25\sqrt{3}x+125)=0$, which has roots $x=-\frac{5\sqrt{3}}{3}$, $x=\frac{5\sqrt{3}}{3}$, $x=-\frac{25}{\sqrt{3}}$, and $x=\frac{25}{\sqrt{3}}$. The intervals can now be determined by testing points within each of the intervals, and seeing whether the second derivative is positive or negative: For $x<-\frac{25\sqrt{3}}{3}$, choose $x=-10$. Then $f''(-10)>0$. For $-\frac{25\sqrt{3}}{3}\frac{25\sqrt{3}}{3}$, choose $x=10$. Then $f''(10)>0$. Therefore, the function is concave upward on $\left(-\infty,-\frac{25\sqrt{3}}{3}\right)$ and $\left(\frac{25\sqrt{3}}{3},\infty\right)$, and concave downward on $\left(-\frac{25\sqrt{3}}{3},\frac{25\sqrt{3}}{3}\right)$.

Determine the Open Intervals on Which the Graph of the Function