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2 2 18 16 14 12 1 graphics 05 05 1 3 now lets

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Unformatted text preview: 1.8 1.6 1.4 1.2 -1 -Graphics- -0.5 0.5 1 3 Now let's compute the second derivative: x x f [x] 5 8x5 + (1+x2 )24x - (1+x2 )2 Log[1+x2 ]3 Log[1+x2 ]2 14x3 (1+x2 )Log[1+x2 ]2 + 6x Log[1+x2 ] Simplify 8x5 (1+x2 )2 Log[1+x2 ]3 + (1+x2 )24x - Log[1+x2 ]2 5 14x3 6x + Log[1+x2 ] (1+x2 )Log[1+x2 ]2 " " 2 2 2 4x5 -x3 (7+5x2 )Log[1+x2 ]+3x(1+x2 ) Log[1+x2 ] And plot it. Since it is zero precisely when x = 0, we see that our critical value of x = 0 happens at an inflection point by the second derivative test. Plot[ " " 2 2 2 4x5 -x3 (7+5x2 )Log[1+x2 ]+3x(1+x2 ) Log[1+x2 ] , (1+x2 )2 Log[1+x2 ]3 {x, -1, 1}] (1+x2 )2 Log[1+x2 ]3 2 1 -1 -0.5 -1 -2 0.5 1 -Graphics- 4...
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