# Determine the intervals on which the function is concave up and intervals on which the function is concave down.f(x)=(x^2)/ (x^2)- 1

Asked on by cspanutius

beckden | High School Teacher | (Level 1) Educator

Posted on

Concave up is when f''(x) > 0 and f''(x) < 0.

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

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

`f''(x) = ((x^2+1)^2(2)-(2x)(2(x^2+1)(2x)))/(x^2+1)^4 = (2(x^2+1)-8x^2)/(x^2+1)^3 = -(2(3x^2-1))/(x^2+1)^3`

Since `(x^2+1)` is always positive, we only have to find when `3x^2-1` is positive.  `3x^2-1` is positve when `x > 1/sqrt(3)` and `x<-1/sqrt(3)` which is the interval `(-1/sqrt(3),1/sqrt(3))` .  So this function is negative on `(-oo,-1/sqrt(3))` and `(1/sqrt(3),-oo)` and positve on `(-1/sqrt(3),1/sqrt(3))` .

The answer is f(x) is concave down on `(-oo,-1/sqrt(3))` and `(1/sqrt(3),oo)` .  The function is concave up on `(-1/sqrt(3),1/sqrt(3))` .   Note that `1/sqrt(3) = sqrt(3)/3`

We can verify from a graph.

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