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 $[-3.497, \infty)$ 
 A: 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 $$-\frac{25}6\sqrt[5]{\frac5{12}}=f\left(\left(5/12\right)^{\,2/5}\right)<0=f(0),$$
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
$$\therefore\boxed{{{\left[{{{{-\frac{25}6\sqrt[5]{\frac5{12}}}};{{\infty}}}}\right)}}}\,.$$
A: Calculus can work in this case.
Consider your function, $$f(x) = 2x^3 - 5\sqrt{x} \ \ \ \ \text{where }x\geq 0$$
Since this function is continuous, the range will be the interval bounded by the minimum and maximum value of $f$ on it's domain.
So we take the derivative, $$f'(x) = 6x^2 - \frac{5}{2\sqrt{x}}$$
Find the values of $x$ such that $f(x)$ is an extrema, that is find values of $x$ where $f'(x) = 0$.
WolframAlpha says the only solution is $x_0 = \frac{(\frac{5}{3})^{\frac{2}{5}}}{2^{\frac{4}{5}}}.$
Take the second derivative, $$f''(x) = \frac{5}{4x^{\frac{3}{2}}} + 12x$$.
Then plug in $x_0$, $$ f''(x_0) = 5\frac{3^{\frac{3}{5}}5^{\frac{2}{5}}}{2^{\frac{4}{5}}} > 0 $$
Since the second derivative is positive at $x_0$, $f(x_0)$ is concave upward at $x_0$, so $f(x_0)$ the minimum bound.
$$f(x_0) \approx -3.497$$
There are no other extrema, so there is no maximum bound on the range of $f$. 
So the range of $f$ is $[-3.497, \infty)$.
A: The obvious answer doesn't work if you're doing pre-calculus, since it involves using calculus. But you can use it as a sanity check still.
Given the relative complexity of doing it the right way, I'm not sure there's a good way to do this without calculus except to use numeric approximations.
Algebra Part
We know there's a minimum graphically, but it's also analytically clear from the equation:
$x^3=\sqrt{x}$ when $x=0,x=1$
$x^3<\sqrt{x}$ when $0<x<1$
$x^3>\sqrt{x}$ when $x>1$
The coefficients (2 and 5) mean the point where the first term is equal to the second term isn't exactly 1 (it's ${5\over 2}^{2\over 5}\approx 1.443$), but the graph does the same thing. Since $2x^3<5\sqrt{x}$ between 0 and 1.443, $y<0$ in that range, and we should find a minimum there.
Since $x^3$ gets positive much faster than $\sqrt{x}$ gets negative, it's analytically clear the function grows without bound to infinity as x gets arbitrarily large.
Calculus Part
$y=2x^3-5\sqrt{x}$
$y'=6x^2-{5\over 2\sqrt{x}}$
$0=6x^2-{5\over 2\sqrt{x}}$
${5\over 2\sqrt{x}}=6x^2$
${5\over 12}=x^2x^{1\over 2}$
$x^{5\over 2}={5\over 12}$
$x={5\over 12}^{2\over 5}=\sqrt[5]{{5\over 12}^2}=\sqrt[5]{25\over 144}\approx 0.608$
This is the x value where your minimum occurs. Plug that into the original equation and to get
$y=2(\sqrt[5]{25\over 144})^3-5\sqrt{\sqrt[5]{25\over 144}}$
$y=2\sqrt[5]{15,625\over 2,985,984}-5\sqrt[10]{25\over 144}$
$y={5\over 6}\sqrt[5]{5\over 12}-5\sqrt[5]{5\over 12}$
$y=({5\over 6}-5)\sqrt[5]{5\over 12}=-{25\over 6}\sqrt[5]{5\over 12}\approx -3.4974$
A: 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!
