Minima of $f(x)=\frac{x^2-1}{x^2+1}$

Question

If $f(x)=\dfrac{x^2-1}{x^2+1}$ for every real $x$ then find the minimum value of $f$

$$ f'(x)=\frac{4x}{(x^2+1)^2}=0\implies x=0\\ f'(-0.5)<0\quad\&\quad f'(0.5)>0 $$ Seems to me like $x=0$ is a point of inflexion. But, $$ f''(x)=\frac{4(1-3x^2)}{(x^2+1)^3}\\ f''(0)>0\implies x=0\text{ is a minima} $$

Am I making some stupid mistake here, since it doesnt make sense to me ?

Answer 1

Your calculations are right. You are just confusing the ideas:

$1)$ $f'(0)=0$ then $x=0$ is a critical point.

$2)$ $f'(-0,5)<0$ and $f'(0.5)>0$ implies that $f$ is decreasing at $-0.5$ and increasing at $0.5$. It reinforce the idea that $x=0$ is a minimum. You confirm that when you see that $f''(0)>0$.

A inflection point occours when the concavity changes. It means the sign of $f''$ changes, not the sign of $f'$.

Answer 2

You may just write

$$\frac{x^2-1}{x^2+1} = 1-\frac{2}{x^2+1}$$

So, the minimum occurs at $x=0$.

Answer 3

Let $t=x^2$, which implies $t\ge0$.

Then, the numerator of the derivative of

$$\frac{t-1}{t+1}$$ is $(t+1)-(t-1)=2$ so that there are no stationary points in $t$ and the function is growing.

Hence, a minimum occurs for $t=x=0$.

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Even simpler,

$$\frac{t-1}{t+1}=1-\frac2{t+1}$$ is obviously growing (as the opposite of the inverse of a growing function).