How do you determine whether the function #f(x)= 4/(x^2+1)# is concave up or concave down and its intervals?

Question

Isabella Ackerman · Accepted Answer

#f(x)# is concave up for #|x| > 1/sqrt5# and concave down for #|x| < 1/sqrt5#

If a function is differentiable twice, we know that it is concave up if #f''(x) >0# and concave down viceversa.

So let us calculate:

#f(x) = 4/(x^2+1) = 4(x^2+1)^(-2)#

#f'(x) = -16x(x^2+1)^(-3)#

#f''(x) = 96x^2(x^2+1)^(-4)-16(x^2+1)^(-3) = frac(96x^2-16(x^2+1)) ((x^2+1)^4)= frac(96x^2-16x^2-16) ((x^2+1)^4)=16frac (5x^2-1) ((x^2+1)^4)#

Evidently the denominator of #f''(x)# is always positive, so the sign of #f''(x)# is the sign of the numerator and we see that:

#f''(x) < 0# for #|x|< 1/sqrt(5)#

So #f(x)# is concave up in the intervals #(-oo,-1/sqrt(5))# and #(1/sqrt(5),+oo)# and concave down for #x# in the interval #(-1/sqrt(5),1/sqrt(5))#.

graph{4/(x^2+1) [-10, 10, -5, 5]}

Emma Aldridge · Answer

undefined To determine the concavity of the function $ f(x) = \frac{4}{x^2 + 1} $ and its intervals, you need to find its second derivative. Then, if the second derivative is positive, the function is concave up, and if it's negative, the function is concave down. The concavity can change at points where the second derivative is zero or undefined. Therefore, to find the intervals of concavity, you can identify the intervals where the second derivative is positive (concave up) or negative (concave down).