A question about graph plotting.

Question

Q. Plot $f(x)=\frac{x^2}{2}-\log(1+x^2)$

My approach:

$$f ' (x) = x\left[1-\frac{2}{(1+x^2)}\right]$$

Equating the above with $0$; I get the maxima and minima points as $0,+1$ and $-1$.

But when I try finding the increasing and decreasing part, say let's find the increasing part, $f'(x)>0$

$$x\left[1-\frac{2}{(1+x^2)}\right] > 0$$

$$x(x^2-1) > 0$$

I am stuck here as the inequality is confusing me. Thus I did not proceed with finding $f''(x)$ to locate distinctively the maxima and minima points among $0,1$ and $-1$ and cannot figure out the graph either...

Please help me solve this.

Thank you.

Answer 1

You need to solve $x(x-1)(x+1)>0$. Consider four intervals: $]-\infty,-1[$, $]-1,0[$, $]0,1[$, $]1,\infty[$. Just take a point from each interval and evaluate the expression to find out whether the expression is positive or negative on that interval.
For example, take $x=-2$ for the first interval: $-2(-2-1)(-2+1)=-2 \cdot (-3) \cdot (-1) <0$ so the expression is negative on $]-\infty,-1[$.

Answer 2

$x(x^2 - 1)$ is a polynomial of third order. Its roots are -1, 0 and 1. You can compute the value in -0.5 and 0.5 to know its sign. You can also notice the positivity of the leading coefficient. This polynomial is positive between -1 and 0, and negative between 0 and 1. Thus, you get the variations of your function between -1 and 1.