where is g(x)=1/(x^2+1) concave down? f(x)=4xe^x find the exact value of x so th

Question

where is g(x)=1/(x^2+1) concave down? f(x)=4xe^x find the exact value of x so that the second derivative is equal to zero derivative of (t^1/2)/t^2+8 i tried doing this and got (t^2+8)(1/2t^-1/2) -t^1/2(2t) but i guess this is wrong

Explanation / Answer

Concave down when g''(x) < 0 Concave up when g''(x) > 0 g(x) = 1 / (x^2 + 1) g'(x) = ( (x^2 + 1) * 0 - 1 * (2x) ) / (x^2 + 1)^2 = -2x / (x^2 + 1)^2 g''(x) = ( (x^2 + 1)^2 * (-2) - (-2x) * (2) * (2x) ) / (x^2 + 1)^4 = (-2 * (x^2 + 1)^2 + 8x^2) / (x^2 + 1)^4 = (8x^2 - 2 * (x^4 + 2x^2 + 1) / (x^2 + 1)^4 = (8x^2 - 2x^4 - 4x^2 - 2) / (x^2 + 1)^4 = (-2x^4 + 4x^2 - 2) / (x^2 + 1)^4 = -(2x^4 - 4x^2 + 2) / (x^2 + 1)^4 = -2 * (x^4 - 2x^2 + 1) / (x^2 + 1)^4 = -2 * (x^2 - 1)^2 / (x^2 + 1)^4 We need to find when this is less than 0. To do this, we first find when it equals 0: -2 * (x^2 - 1)^2 = 0 x^2 - 1 = 0 x^2 = 1 x = +/- 1 This gives us 3 domains -infinity

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