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Math Help - Find the intervals on which f(x) = ln(x + 1) is concave up

  1. #1
    VkL
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    Find the intervals on which f(x) = ln(x + 1) is concave up

    Find the intervals on which f(x) = ln(x + 1) is concave up

    I got critical points to be -1, and 1

    I checked by the first derivative that from - infinity to 0 the slope is negative and from 0 to positive infinity the slope is positive, i get confused about writing the intervals.
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  2. #2
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    Quote Originally Posted by VkL View Post
    Find the intervals on which f(x) = ln(x + 1) is concave up

    I got critical points to be -1, and 1

    I checked by the first derivative that from - infinity to 0 the slope is negative and from 0 to positive infinity the slope is positive, i get confused about writing the intervals.
    We need to find the second derivative. First f'(x) = \tfrac{(x^2+1)'}{x^2+1} = \tfrac{2x}{x^2+1} by the chain rule.
    Second, f''(x) = \tfrac{(2x)'(x^2+1) - (2x)(x^2+1)'}{(x^2+1)^2} = \tfrac{2(1-x^2)}{(x^2+1)^2} . It concaves up when f''(x) > 0 so \tfrac{2(1-x^2)}{(x^2+1)^2} > 0 . Multiply this inequality both sides by \tfrac{(x^2+1)^2}{2} to get: \tfrac{(x^2+1)^2}{2} \cdot \tfrac{2(1-x^2)}{(x^2+1)^2}  > 0 \cdot \tfrac{(x^2+1)^2}{2}*. Thus, 1-x^2 > 0 \implies x^2 < 1 \implies -1 < x < 1.


    *)Remember that \tfrac{(x^2+1)^2}{2} > 0 so the inequality does not flip.
    ----

    There is another way to do this problem, a more straightforward way that you probably learned at college but a little longer. First, solve f''(x) = 0 and that happens when \tfrac{2(1-x^2)}{(x^2+1)^2} = 0 \implies 1-x^2 = 0 \implies x = \pm 1. Now put the points x=-1, x=1 on the number line. Take any point to the left of -1, say -2 and compute f''(-2) < 0 ---> thus f concaves down for -\infty < x < -1. Take any point between -1 and 1, say 0 and compute f''(0) > 0 ---> thus f concave up for -1 < x < 1. Take any point to the right of 1, say 2 and compute f''(2) < 0 ---> thus f concaves down for 1<x<\infty.
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