Number of zeros elementary functions While studying functions and zeros of functions i came across the following problem:
Problem:
Knowing that $1,09<\ln 3<1,1,$ determine the number of zeros of $J: \mathbb{R} \rightarrow \mathbb{R}$ given by $J(x)=3^{x}-1-2 x^{2}$
Attempt:
One quickly notices that $x = 0$ satisfies the expression given in the problem. Now, one can draw the functions $3^{x}$ and $2x^2 + 1$ and notice that for $x < 0$ is not possible  that $3^x = 1 + 2x^2$
Therefore, we should investigate what happens when $x > 0$. For that, i tried solving $$
g^{\prime}(x)>h^{\prime}(x)
$$
where $g(x) = 3^x$ and $h(x) = 1 + 2x^2$. The ideia behind solving the above inequality comes from the fact that if $g^{\prime}(x)>h^{\prime}(x)$ for every $x > 0$, then it would not possible for $g(x) = h(x)$ for $x > 0$ (Just intuition, i dont know if that holds in general)
However, $$
g^{\prime}(x)>h^{\prime}(x) \Rightarrow \ln (3) \cdot 3^{x}>4 x+1
$$
It looks like we are on the right track, since the term $\ln(3)$ just appeared. Now, by observing that $$
1,1 \cdot 3^{x}> ln(3) \cdot 3^x> 4 x+1>4 x
$$
we see that $$
\frac{1,1 \cdot 3^{x}}{x}>4
$$ which i dont know how to solve.
Can someone help?
Thanks in advance, Lucas

In fact, it is easy to verify that $x = 0, 1, 2$ are all zeros of $J$.
We can show that these are the only zeros of $J$, in the following order:

*

*Show that $J''$ is strictly increasing, and deduce that $J''$ has exactly one zero, which lies in the interval $(0, 2)$. We call this zero $z$.


*Show that $J'$ is strictly decreasing on the interval $(-\infty, z)$ and strictly increasing on the interval $(z, \infty)$. Deduce that $J'$ has exactly two zeros, which lie in the intervals $(0, 1)$ and $(1, 2)$, respectively. We call them $z_1$ and $z_2$.


*Show that $J$ is strictly increasing on the intervals $(-\infty, z_1)$ and $(z_2, \infty)$ and is strictly decreasing on the interval $(z_1, z_2)$. Thus $J$ has at most three zeros, which we have all found.
