Absolute minimum of $f(x) = \sqrt{2x^2-3x+4} + \sqrt{x^2-2x}$ I've got the domain of function and I've attempted to find the first derivative at zero but it results in a quartic equation that is too difficult for me to solve. 
$f'(x) = \frac{4x-3}{\sqrt{2x^2-3x+4}} + \frac{2x-2}{\sqrt{x^2-2x}}$
For $f'(x) = 0$:
$(4x-3)^2(x^2-2x) = (2-2x)^2(2x^2-3x+4)$
Therefore $8x^4-28x^3+9x^2+26x-16 = 0$
I've probably made an error in my calculations but I'm sure that this is not how you approach the question and I'm not quite sure how to do it otherwise. According to the textbook the answer is 2.
 A: Let $f(x)=\sqrt{2x^2-3x+4}$, $g(x)=\sqrt{x^2-2x}$ and $h(x)=\sqrt{x}$.
Since $h$ is an increasing function, 
we see that $f$ and $g$ are decreasing functions on $(-\infty,0]$
and $f$ and $g$ are increasing functions on $[2,+\infty)$.
Thus, $$\min\limits_{(-\infty,0]\cup[2,+\infty)}\left(\sqrt{2x^2-3x+4}+ \sqrt{x^2-2x}\right)=\min\{(f+g)(0),(f+g)(2)\}=2$$
A: First of all, you have a slight error in your derivative: you're missing a factor of $\frac{1}{2}$ in both terms of $f'$. Fortunately, it doesn't affect solving the equation $f'(x)=0$.
You said that you found the domain of this function, so you know that the domain is $(-\infty,0]\cup[2,+\infty)$.
So we need to solve the following:
$$f'(x)=\frac{4x-3}{2\sqrt{2x^2-3x+4}}+\frac{2x-2}{2\sqrt{x^2-2x}}=0.$$
The domain of the derivative is $(-\infty,0)\cup(2,+\infty)$, and it's fairly easy to see that no point in this domain can satisfy the equation: if $x>2$, then both terms of $f'$ are positive, and if $x<0$, then both terms of $f'$ are negative.
However, we also have two more critical points determined by the condition that $f'(x)$ DNE, which happen to be the endpoints $x=0$ and $x=2$ of the domain. So extreme values can only occur at these two points. Both give us endpoint minima (we can conclude that from the signs of $f'$ determined above). Plugging them into the original function yields the answer.
SOME EXTRA EXPLANATION
For the sake of completeness of this answer, let's see what happens if we actually try to solve the equation. Multiplying by $2$ and rearranging terms, we get:
$$\frac{4x-3}{\sqrt{2x^2-3x+4}}=-\frac{2x-2}{\sqrt{x^2-2x}}.$$
Now we've got to be careful! We're going to square both sides, which may create extraneous roots, so don't forget to verify the roots afterwards. Squaring and applying the "cross-multiply" property indeed produces the equation
$$8x^4-28x^3+9x^2+26x-16=0.$$
According to Wolfram Mathematica, this equation has only two real roots $x\approx-0.97$ and $x\approx2.76$.  But substituting them back into $f'(x)=0$ shows that they do not satisfy the original equation — they both happen to be such extraneous "roots" introduced by squaring. 
A: We know that $x\in(-\infty,0)or(2,+\infty)$
We can draw the image of the $(x^{2}-2x)^{\frac{1}{2}}$and $(2x^{2}-3x+4)^{\frac{1}{2}}$,you will find the min of this function is 2 when x=0;
A: I would suggest sketching your first square root and your second square root separately on a graph. The solution then doesn't require any sort of complicated algebra.

