Determinate analytically the range of a given function Is there a general way to obtain analytically the range of a function?

For instance, for the function $f(x)=4+\sqrt{x-3}$, I proceed as follows:

If $y \in ran(f)$ then for some $x$, $4+\sqrt{x-3} = y$ then $y-4 = \sqrt{x-3}$ now since we only consider non negative values of the square root then $y-4 = \sqrt{x-3} \geq 0$ so $y \geq 4$ and $ran(f) = [4, \infty)$. 

But for the function $f(x)= 2x^{3}-5\sqrt{x}$  I can't seem to find a similar procedure to obtain the range analytically, of course I can plot the function and I get that the range is $[-3.497, \infty)$
Since $f:\mathbb R\mapsto\mathbb R$, $x\geq0$ for $\sqrt x$ to be real. Also,
$$\lim_{x\to\infty}f(x)=\infty$$
and
$$f(0)=0,$$
therefore $f$ is bounded from below since it is continuous on ${({{0};{\infty}})}$.
$$f'(x)=\frac{\mathrm d}{\mathrm dx}f(x)=6x^2-\frac5{2\sqrt x}=\frac1{\sqrt x}\left(6x^{5/2}-\frac52\right);$$
hence the minima are given by
$$f'(x)=0\iff6x^{5/2}-\frac52=0\iff x^{5/2}=\frac5{12}\iff x=\left(5/12\right)^{\,2/5};$$
so the lower bound is either at $x=0$ or $x=\left(5/12\right)^{\,2/5}$;
$$f\left(\left(5/12\right)^{\,2/5}\right)=2\left(5/12\right)^{\,6/5}-5\left(5/12\right)^{\,1/5}=\frac{5}{6}\sqrt[5]{\frac5{12}}-5\sqrt[5]{\frac5{12}}=-\frac{25}6\sqrt[5]{\frac5{12}};$$
since $$-\frac{25}6\sqrt[5]{\frac5{12}}=f\left(\left(5/12\right)^{\,2/5}\right)&lt;0=f(0),$$
the minimum at $\left(\left(5/12\right)^{\,2/5}\right)$ provides the lower bound, that is, the image of $f:\,{{[{{{{0}};{{\infty}}}})}}\mapsto\mathbb R$ such that $f(x)=2x^3-5\sqrt x$ is
$$\therefore\boxed{{{\left[{{{{-\frac{25}6\sqrt[5]{\frac5{12}}}};{{\infty}}}}\right)}}}\,.$$
Domain: $x\ge 0$
Slopes: $f'(x)=6x^2-\frac52x^{-\frac12}$.

Now what does it indicate when the slope is positive/negative/zero? 

Finally, plot the function (loosely, but it's gonna be sufficiently informative about your problem) by hand!