Two Steady Increasing Functions

Make a substitution $t=2^{x^2}+1$ with $t\geq 2$ since $x^2\geq0$ . Now $f(t)=t-2+\frac{2}{t}$ is increasing for $t\geq 2$ since $f'(t)=1-\frac{2}{t^2}\geq \frac{1}{2}$ , so that the minimum of $f(t)$ on the interval $[2,\infty)$ is $f(2)=\boxed{1}$

$f(x)=2^{x^2}-1+\dfrac{2}{2^{x^2}+1}$ $f'(x)=2x(2^{x^2})(\ln 2)-\dfrac{2}{(2^{x^2}+1)^2}(2x(2^{x^2})(\ln 2))$ $=2x\color{#0C6AC7}{\left((2^{x^2})(\ln 2)\left(\dfrac{(2^{x^2}+1)^2-2}{(2^{x^2}+1)^2}\right)\right)}$ We can see term written in blue is always positive therefore sign of $f'(x)$ is governed by the sign as $x$ . $\implies f'(x)>0 \forall x>0\implies \text{f(x) is increasing in} (0,\infty)\\ \text{and} ~f'(x)<0 \forall x<0\implies \text{f(x) is decreasing in}(-\infty,0)$ And therefore has a global minima at $x=0$ . $f(0)=\boxed 1$ Which is the required minimum value of $f(x)$

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$\mathbf{NOTE:-}$ $\mathbf{(1).}~x^2>0\implies 2^{x^2}>1 \implies 2^{x^2}+1>2.$ And hence blue term is always positive.

Here's a way I first approached the problem and got it wrong.. :-) $\mathbf{(2).} \color{#456461}{\left(2^{x^{2}}+1+\frac{2}{2^{x^{2}}+1}\right)}-2$ Apply $AM-GM$ on coloured terms to get minimum value as $2(\sqrt 2-1)$ but that's wrong since case for equality gives: ${2^{x^{2}}+1}=\sqrt 2$ While we previously noted that ${2^{x^{2}}+1}>2$ so $AM-GM$ could not be applied this way and hence gives wrong result..

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P.S:- Graphing calculator was also a option but I wanted to write a solution involving pure calculus..