Total no. of real solution of $\displaystyle 2^x = 1-x^2$

My Trail:: Let $\displaystyle f(x) = 2^x$ and $\displaystyle g(x) = 1-x^2$

Now Using first derivative test, we will check whose slope is Increasing faster rate.

So $\displaystyle f{'}(x) = 2^x\cdot \ln (2)$ and $\displaystyle g^{'}(x) = -2x$

So for The $\displaystyle x\leq -1$, Slope of $\displaystyle f(x) > $ Slope of $\displaystyle g(x)$

So $\displaystyle f(x)$ is always above $\displaystyle g(x)$

So no solution for $\displaystyle x\leq -1$

and for $\displaystyle x \geq 0,$ Slope of $\displaystyle f(x) > $Slope of $\displaystyle g(x)$

So $\displaystyle f(x)$ is always above $\displaystyle g(x)$

So no solution for $\displaystyle x > 0$

Now we will calculate in $\displaystyle -1 < x < 0$

Now I Did not understand How can I find whether $\displaystyle f(x)$ is $\displaystyle g(x)$ or not

please help me

Thanks