# 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.

• Show us your work Jan 8 '17 at 4:41
• Hint: determine the domain of $f$ first. Then, if $f'$ has no zeros in that domain, look at the boundary points for possible extrema.
– dxiv
Jan 8 '17 at 5:00

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, $$\min\limits_{(-\infty,0]\cup[2,+\infty)}\left(\sqrt{2x^2-3x+4}+ \sqrt{x^2-2x}\right)=\min\{(f+g)(0),(f+g)(2)\}=2$$

• What's the role of $h$ Jan 8 '17 at 5:19
• @MyGlasses $h$ is an increasing function. Thus, if $f$ is an increasing so $h\circ f$ is an increasing. If $f$ is a decreasing so $h\circ f$ is a decreasing. Here $(h\circ f)(x)=h(f(x))$. Jan 8 '17 at 5:24
• Do you mean $f_1=2x^2-3x+4$ and $f=hof_1$~? Jan 8 '17 at 5:27
• @MyGlasses Yes, of course! But in my previous reasoning $f$ is something another. Jan 8 '17 at 5:29

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.

• Is there a definitive way you can prove that there is not point satisfying the equation, in terms of working out. Jan 8 '17 at 5:35
• I did: in the paragraph that starts with The domain of the derivative is... Jan 8 '17 at 5:38
• No i was just wondering how would you write it mathematically. e.g. If x<0, f(x)<0 Jan 8 '17 at 5:50
• Something like this. First of all, for all $x$ in the domain, both denominators, being square roots whose domains are already taken into account, are positive. If $x>2$, then $4x-3>0$, so the first fraction is $>0$; and $2x-2>0$, so the second fraction is also $>0$; therefore for $x>2$, $f'(x)>0$ as the sum of two positive numbers. Similarly for $x<0$. Jan 8 '17 at 5:57

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;

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. • If anyone would explain the downvotes I'll happily elaborate.
– Matt
Jan 8 '17 at 5:14
• Do not take it seriously, there are many fools here. They simply are just some mean people! Jan 8 '17 at 5:18
• any "fool" can get a graphing calculator and find the answer. You need to actually solve the problem. Jan 8 '17 at 5:22
• dezdichado, he just simply put his answer here. That is all about how he thinks the problem. He does not have the responsibility to make sure his solution is right, or perfect in anybody's eyes. If you can solve the problem, then you put your answer here. Please, do not judge other people that easily. Jan 8 '17 at 5:26
• I didn't actually suggest using a graphing calculator or say OP needed one. I graphed it using online software to illustrate what I meant - this is not a difficult concept to understand. I could alternatively have sketched it on graph paper, taken a picture, and uploaded that. That is not preferable. It is not hard to understand my answer to this question or why it is useful to have the plot. I don't understand your attitude, nor your downvote. But thanks, I suppose.
– Matt
Jan 8 '17 at 5:29