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Determine the intervals on which the function is concave up and intervals on which the...
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High School Teacher
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.
Posted by beckden on July 18, 2012 at 1:17 AM (Answer #1)
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