Question regarding these two extreme values problems...

**theant4** · May 11th 2011, 03:42 PM

So I'm on the last two problems of this assignment and I've worked them both out as far as I can and I can't seem to figure out where I'm going wrong. Both of them have to do with extreme values and the 1st and 2nd Derivative Tests.

Here's the first:
$f(x) = \cos (x) + \frac{\sqrt{2}}{2} x$

The function has two critical numbers, A being less than B, over the interval $[0, 2\pi ]$
A = ( ? )
B = ( ? )
f''(A) = ( ? )
f''(B) = ( ? )
Thus f(x) has a local ( ? ) at A and a local ( ? ) at B.

To start out, I got this for the derivative:
$f'(x) = -\sin (x) + \frac{1}{\sqrt{2}}$
$0 = -\sin (x) + \frac{1}{\sqrt{2}}$
$\frac{-1}{\sqrt{2}} = -\sin (x)$
$\frac{1}{\sqrt{2}} = \sin (x)$

And now I have to solve for x and I'm not really sure how. I took the second derivative and got $f''(x) = -\cos (x)$ . I don't feel like I did the problem right at all... where am I going wrong? Where do I go next?

Here's the second problem:
$f(x) = x(\sqrt{{x}^{2} + 4})$ on the interval $[-5,4]$

-f(x) is concave down on the region ( ? ) to ( ? )
-f(x) is concave up on the region ( ? ) to ( ? )
-The inflection point for the function is at ( ? )
-The minimum for the function occurs at ( ? )
-The maximum for the function occurs at ( ? )

Again I started by taking the derivative:
$f'(x) = \frac{2{x}^{2} + 4}{ \sqrt{{x}^{2} + 4}}$

Now, that $\sqrt{{x}^{2} + 4}$ should just simplify to x + 2 right? In that case, I continued:
$\frac{2{x}^{2} + 4}{x + 2}$
One of the x's cancels...
$\frac{2x + 4}{2}$
And if I keep going and solve for x I find that x = 2. The only problem is that the question is asking for multiple intervals on which the function is increasing and decreasing and with this single critical point it's impossible to find those intervals. I know I'm definitely doing something wrong in this problem.

Any tips on either of these would be greatly appreciated...

P.S. And also how do I get rid of the </br> for every time I use the Latex thing? Moderator Edit: Its weird, it inserted the html command for breaks <br /> if you create a new line after declaring the first [tex] bracket and hit enter before the second [tex] bracket. I'll look into this more...

**TKHunny** · May 11th 2011, 05:37 PM

Originally Posted by theant4

$\frac{1}{\sqrt{2}} = \sin (x)$

And now I have to solve for x and I'm not really sure how.

You're serious? Somewhere, sometime, you should have memorized $\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}$

There's another one, too.

Still further, the derivative doesn't know about your endpoints. Check out $x = 0$ and $x = 2\pi$

**mathprofessor** · May 11th 2011, 06:34 PM

(2X^2 +4)/(x+2). You are NOT dividing 2x^2 by x, but rather by x+2. The x's do not cancel out. Consider this simple example:
(2*5+2)/(2+3) which is (10+2) /5 =12/5. You will say that since the 2's cancel out then (2*5+2)/(2+3)=(5+2)/3 =7/3 or maybe you'd say (5+2)/(1+3)=7/4. Both are wrong since you are dividing 2*5 by (2+3) not by 2!! You need to learn the difference between factors (which can be canceled out) and terms (which can not be canceled out usually).
Steven

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