How many zeroes does the function $ f(x)=\exp(x)-3x^2$ have in $\mathbb{R}$? My try:
Attempt (1):  $f(x)=0$, which gives $\exp(x)=3x^2$.
Since $\exp(x)$ and $3x^2$ intersect at exactly two points, therefore the function has two zeros.
Attempt (2): I traced $f(x)=\exp(x)-3x^2$ which cuts the $X$-axis at three points, therefore $f(x)$ has three zeros.
Which one is wrong? I need help.
 A: The function $f(x)=\exp x-3x^2$ has
$$
\lim_{x\to-\infty}f(x)=-\infty \qquad \lim_{x\to\infty}f(x)=\infty
$$
Moreover, $f'(x)=\exp x-6x$. We want to see where the derivative vanishes, so we compute $f''(x)=\exp x-6$; this shows $f'$ has a minimum at $\log6$ and
$$
f'(\log 6)=6-6\log6=6(1-\log 6)<0
$$
Since $\lim_{x\to-\infty}f'(x)=\infty=\lim_{x\to\infty}f'(x)$ we conclude $f'$ vanishes twice, say at points $\alpha$ and $\beta$, with $\alpha<\beta$; now
$$
f(\alpha)=\exp\alpha-3\alpha^2=6\alpha-3\alpha^2=3\alpha(2-\alpha)
$$
and, similarly,
$$
f(\beta)=3\beta(2-\beta)
$$
Note that $f'(0)=1$ and $f'(2)=e^2-12<0$; therefore $0<\alpha<2$ and $\beta>2$.
Thus the local maximum $f(\alpha)>0$ and the local minimum $f(\beta)<0$, so the equation has three solutions.
Note that $y=\exp x$ and $y=3x^2$ actually intersect in three points, although it's not easy to see it from a hand-made graph.
You can see the three intersection in the following graph, where I used $x\to x/\sqrt{3}$ for making it “smaller”.

A: Look at the second derivative $e^x-6$ it has one real solution and so the first derivative has two solutions and the function has three real solutions
A: We have
$$f(x)=e^x-3x^2$$
so
$$f'(x)=e^x-6x$$
and
$$f''(x)=e^x-6$$
This means that $f''(x)>0$ for $x>\ln6$ and $f''(x)<0$ for $x<\ln6$.  We know that $\ln6\in(1,2)$. 
We can use intermediate value theorem to see that $f(\alpha)=0$ for some $\alpha\in(0,1).$ Hence that turning point is a maximum.
We can again use intermediate value theorem to see that $f(\beta)=0$ for some $\alpha\in(2,3).$ Hence that turning point is a minimum.
This means that for $x<\alpha$, $f$ is increasing. For $\alpha<x<\beta$, $f$ is decreasing. For $x>\beta$, $f$ is increasing.
Now that we found three regions where $f$ is monotone inside each, we can apply intermediate value theorem (basically brute force) and be sure that each time the product is negative, that range only has one root.
Eventually, you should be able to confirm that there are three real roots in total.
A: Following up on Sonnhard Graubner's comment:
Take the square root of $e^x=3x^2$ and from  $x=\pm\sqrt3 e^{x/2}$ establish one side as $ue^u$ using $u=-\frac x2$, $$-\frac x2e^{-\frac x2}=\pm\frac1{2\sqrt3},$$ so that finally one can write down the roots as $$x_1=-2W_0(\frac1{2\sqrt3}),\ x_2=-2W_0(-\frac1{2\sqrt3}),\ x_3=-2W_{-1}(-\frac1{2\sqrt3})$$ using the Lambert-W function which has one branch $W_0$ for positive arguments and two branches $W_0,W_{-1}$ for arguments in $(-e^{-1},0)$, which this argument falls into as $2\sqrt3=\sqrt{12}>3>e$.
A: We have $e^x=3x^2\iff x=\log(3x^2)=\log3+2\log|x|$. (Note, in particular, the absolute value signage!). If you sketch the (concave down) graph of $y=\log3+2\log|x|$ carefully enough (noting that $\log3\gt\log e=1$), you can see that it intersects the straight line $y=x$ at exactly three points.
