# Thread: Find the intervals on which f(x) = ln(x² + 1) is concave up

1. ## Find the intervals on which f(x) = ln(x² + 1) is concave up

Find the intervals on which f(x) = ln(x² + 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.

2. Originally Posted by VkL
Find the intervals on which f(x) = ln(x² + 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.
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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.

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# interval ln x

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