<p>How do you find the critical numbers and the extrema of this function:</p>
<pre><code> 2x((x+1)^1/2)
</code></pre>
<p>without graphing and using a ti83?</p>
<p>thanks for the help.</p>
<p>This should be moved to High School Life or maybe SAT prep, but it shouldn't be here in College Admissions</p>
<p>you find the derivative and set equal to 0.</p>
<p>those are your critical points, remember to includ the endpoints, if applicable.</p>
<p>Then, you take the critical points and do the 1st derivative test.</p>
<p>I assume he/she is having a hard time finding the derivative...</p>
<p>First of all, the domain is [-1,infinity) b/c you can't have square roots of negative numbers</p>
<p>f(x) = 2x(x+1)^1/2</p>
<p>u=2x
du= 2</p>
<p>dv = (x+1)^1/2
v = (2/3)(x+1)^(3/2)</p>
<p>Therefore,</p>
<p>(4/3)(x)(x+1)^3/2 + integral[(4/3)(x+1)^3/2]</p>
<p>(20/15)(x)(x+1)^3/2 + (8/15)(x+1)^5/2 = 0</p>
<p>(4/15)(x+1)^3/2 [5x + 2x + 2] = 0</p>
<p>Critical Values are when:</p>
<p>(x+1)^3/2 = 0 ............. -1
7x + 2 = 0 .............. -2/7</p>
<p>(-infin, -1) DNE
(-1, -2/7) (+)(-) = -
(-2/7, infin) (+)(+) = +</p>
<p>Therefore, critical values are -1 & -2/7. Relative min at -2/7. Check for absolute max/min by plugging into original.</p>
<p>shouldn't dv=(1/2)(x+1)^-1/2?</p>
<p>^ No.
But I don't get why you are integrating.
The critical points are where they derivative of the function is equal to 0, and extrema are the points where there exists a local or absolute maximum/minimum. Don't forget endpoints. So, why are you finding the integral?</p>
<p>can someone please help!!</p>
<p>My problem is that I don't know how to solve for 0.</p>
<p>LOL I am so burned out of calc it is unbelievable...final last week. Oh that is a really stupid question...product rule. lol</p>
<p>f'(x) = 2(x+1)^1/2 + 2x(x+1)^-1/2 = 0</p>
<p>2(x+1)^-1/2 * [ (x+1) + x) = 0</p>
<p>the first part cancels out thru cross multiply...</p>
<p>2 (2x + 1) = 0</p>
<p>x = -1/2 </p>
<p>Min @ x = -1/2. Not sure if you'd consider x = -1 a critical point though; don't believe so...</p>
<p>the answers in the back of my book aren't the answers you have, stambiark.</p>
<p>Is that any better? Plus you have to include the absolute max/min by plugging in the endpoints.</p>
<p>fuc it, you won't be using this in life......ever.</p>
<p>stambliark, can you solve the 0 for this:</p>
<p>2(x+1)^1/2+x(x+1)^-1/2</p>
<p>this'll be my final post....thank you so much and may God bless your soul, Stambliark.</p>
<p>f(x)= 2x((x+1)^1/2)
f'(x)= 2(x+1)^(1/2) + x[(x+1)^(-1/2)
f'(x)=0 if 3x+2/[(x+10)^1/2] = 0 and we know that the denominator can't be equal to zero. So x= -2/3
Df= [-1; 00] so you plug in -1 and -2/3 to find the maxima and the minima.</p>
<p>berthran is right. My god am I dumb. :)</p>
<p>I know this is a really silly question compared to what was asked earlier, but is 0 considered an even number?</p>
<p>Different teachers with different responses to that question I asked many times. They've never said that it was an odd number though. So, not odd. Some say that it's neither odd, nor even.
This is my own idea. If it is located between two odd numbers (-1 and 1,) then it is an even number.</p>
<p>Thanks for the input, berthran. </p>
<p>This question arose because I was taking a practice test from the Kaplan 2005 PSAT Prep book, and one of the math question's explanations stated that 0 was an even number. Hopefully, they won't have debatable questions like that on the test. By the way, I would not really recommend buying the Kaplan prep for SAT because when I took the free SAT given by the College Board, I scored a 2250, but a day later I took one made by Kaplan and scored 2150. Just shows that the questions are inaccurate or the scoring system is messed up.</p>
<p>i tried using the kaplan book for sat but it was a bit too much</p>
<p>if you were me for all things go with the princeton review</p>
<p>just a side note, cooljoe, the reason you scored lower on the kaplan test is because they make their tests more difficult, throw in more curveball questions to prepare you for the real thing</p>
<p>i hate kaplannnnnn</p>