# Given $f'(x)=(2x/2\sqrt{x^2+2})-3$, what are the critical points?

I've been given a function $$y=\sqrt{2+x^2}-3x$$

and I need to find the absolute minimum and maximum between $[5,6]$. I've already found (assuming I did it right) the derivative of y to be $$f'(x)=\frac{2x}{2\sqrt{x^2+2}}-3$$

Now I need to find the critical values, but I'm not sure if I did something wrong or if I don't know how to do it given this problem. I've come up to a roadblock because I'm dealing with a square root, and I can't get all the x variables to one side, aside from having $$\frac{2x}{2\sqrt{x^2+2}}=3$$ or $$x=3\sqrt{x^2+2}$$

Can somebody point me in the right direction? Thanks.

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it mut be "minus 3" in the derivative... – DonAntonio Nov 23 '12 at 1:06
Sorry - I had it written down right but copied it incorrectly here. I fixed it. – agent154 Nov 23 '12 at 1:12
You also lost the factor of 3, as in $x = 3\sqrt{x^2+2}$ – amWhy Nov 23 '12 at 1:15
You should have x = 3 sqrt[x^2 + 2], which should lead to x = (3/2)i (imaginary result). Can you find your errors (for example the 2 in the numerator and the one outside the sqrt should not be there). So, what if you put the 3 back in, square both sides and solve for x? Is that enough of a hint? Of course, what does an imaginary result mean in this context? – Amzoti Nov 23 '12 at 1:15
Simply square both sides, as you thought to do, but now with the corrected derivation... – amWhy Nov 23 '12 at 1:17

After a little manipulation, we reach $$x=3\sqrt{2+x^2}.$$ A reasonable thing to do is to square both sides. We reach $x^2=18+9x^2$.

It is easy to see that this equation has no real solution. so there are no critical points.

Thus the max and min occur at the endpoints. Substitute $5$ and $6$ to find out which is which. Actually, since $f'(x)$ is negative, we know the max is at $5$ and the min is at $6$.

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It's easy to see the function's derivative doesn't vanish where you want (in fact, it vanishes nowhere). Most probably what was meant is to check the extreme points of the given interval, as they are always extreme points of a function defined on a finite interval per definition, thus:

$$f(5)=\sqrt{2+25}-3\cdot 5=\sqrt{27}-15$$

$$f(6)=\sqrt{2+36}-3\cdot 6=\sqrt{38}-18$$

and since $\,f'(x)<0\,$, we get that $\,(5,f(5))\,$ is a local maximum, and $\,(6,f(6))\,$ is a local minimum. Please note that the extreme points of a function in a finite interval are always one-sided local extreme.

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