<p>1) lim (x>0) (sinxcosx)/x</p>
<p>2) Let h be a differentiable function, and let f be the function defined by f(x) = h(x^2 - 3). Which of the following is equal to f'(2)?
A) h' (1)
B) 4h' (1)
C) 4h' (2)
D) h' (4)
E) 4h' (4)</p>
<p>3) Let f be the function defined above, where c and d are constants. If f is differentiable at x = 2, what is the value of c + d?
f(x) = {cx + d for x less than or equal to 2]
{x^2 - cx for x>2}</p>
<p>1) lim(x->0)(sinx/x)=0 I think you can multiply limits?</p>
<p>The answer to that one is 1 according to my calculator but I’m not sure how. I tried l’hospital but it doesn’t seem to work out</p>
<p>for number 1, it’s one. According to one of the limit formula. lim (x>0) sin x/x =1 and cos (0) = 1. So, lim (x>0) (sinxcosx)/x = lim (x>0) (sinx/x) (cos x) = (1)(1)=1</p>
<p>^ yep yep</p>
<p>I finally figure out 3. Since the function is differentiable at x=2, both step functions have the same f(x) when you plug in x=2. The only way to solve for c is to differentiate both step functions. Set the derivative functions equal, since same f(x) means same f’(x) at that point, and you have c=2x-c. Plug in x=2 because that is where the two derivatives are equal. Solve for c and you get c=2. </p>
<p>Plug in c=2 to the step functions and you get 2x+d and x^2-2x. 2x+d=x^2-2x cuz differentiable at x=2, meaning same point. Plug in 2 and solve for d. d=-4</p>
<p>c+d=-2.</p>
<p>Finally figure out 2. But I was being hypothetical in this problem. If f(x)=h(x^2-3), let make the two functions equal to something else. For example, let’s go with this f(x)=x=h(x^2-3).
Substitute the x. So, f(x)=x^2-3=h(x). Differentiate. f’(x)=2x=h’(x). f’(2)=4=h’(2). Yeah, h’(2) isn’t one of the answer. So, that’s why you go to x=h(x^2-3). Differentiate it. 1=h’(x^2-3). x^2-3 would be 1. So, 1=h’(1). And f’(2)=4h’(1)</p>
<p>So it’s B. but i’m kinda unsure.</p>
<p>Thanks for your help jerrry! I understand #3 but I’m not really following what you said for #2. Can someone perhaps, help clarify or offer a alternate way to solve the problem?</p>
<p>According to the chain rule: f’(x) = h’(g(x))<em>g’(x)
for this problem I’m gonna let g(x) = x^2-3
So, f’(x) = h’(x^2 - 3)</em>2x, now it wants specifically at x = 2, so we plug in 2
f’(2) = h’(2^2 - 3)<em>2</em>2 or f’(2) = 4h’(1)</p>
<p>Do you understand?</p>
<p>Oh! I get it now. Thanks guys.</p>
<p>Actually can someone help me with these last two?
<a href=“http://i114.■■■■■■■■■■■■■■■/albums/n251/SignificaFire/CCI02212010_00000.jpg[/url]”>http://i114.■■■■■■■■■■■■■■■/albums/n251/SignificaFire/CCI02212010_00000.jpg</a></p>
<p>I think the second one is B from using my calculator but its part of the non calculator section.</p>
<p>I think that you need to integrate the differential equations given and then compare the antiderivative to the graph given (i.e. find the one with a horizontal asymptote of y=200 and P(0) is about 30). I’m not sure how you could do that without a calculator, though.</p>
<p>As to the 1st one, I have no clue.</p>
<p>For the first one by reverse product rule:
int (f’(x)g(x) + f(x)g’(x)) = f(x)g(x)
int (f(x)g’(x)) = f(x)g(x) - int(f’(x)g(x))
now just apply this to definite integrals.</p>
<p>for the second question. I got A (using a calculator). After integrating dP/dt=0.2P-0.001P^2, I got P=t(.2P-.001P^2). Then I solve for P, to have P on one side and t on the other side, and the function is P=[200(t-0.5)]/t. And this function is the answer because 200t/t=horizontal asymptote.</p>
<p>The first one is integration by parts: Integral of (udv) = uv - Integral of (vdu)</p>
<p>Oh, for second question, I just realize you don’t need a calculator. Since P=200, that is where dP/dt=0. So, just plug in 200 and find the equation where dP/dt=0. And it’s A.</p>
<p>Ugh sorry for the late response, have been pretty busy lately</p>
<p>Anyway for the first one, I got to
5 = f(x)g(x) - int(f(x)g’(x))
This was using vdu, but how do you figure out f(x)g(x)?</p>
<p>Also last one, I promise if anyone doesn’t mind
The polynomial function f has selected values of its second derivative f"" given in the table above. Which of the following statements must be true?
Chart
x, f ‘’(x)
(0,5)
(1,0)
(2,-7)
(3,4)</p>
<p>I’m stuck between D and E
A) f is increasing on the interval (0,2)
B) f is decreasing on the interval (0,2)
C) f has a local maximum at x = 1
D) The graph of f has a point of inflection at x = 1
E) The graph of f changes concavity in the interval (0,2)</p>
<p>I’m leaning towards D cause its (0,2) instead of [0,2], let me know what you guys think</p>
<p>Integral of f(x)g’(x) = f(x)g(x) - int[g(x)f’(x)]</p>
<p>They gave you what the int[g(x)f’(x)] equals from 0 to 1, that is five, so the problem becomes f(x)g(x) from 0 to 1, minus 5. So it will be this:
[f(1)g(1) - f(0)g(0)] - 5, that gives you the answer</p>
<p>Can the question have multiple answers? If the graph has an inflection point at x = 1, which the chart indicates because the second derivative equals 0, that means that at that point it changes concavity, so if D is correct then I would think that E has to be correct as well.</p>
<p>Nope, its definitely one answer. That’s why I’m stuck between D and E.</p>
<p>Thanks for the help on the other problems btw, forgot to put f(x)g(x) from 1 to 0</p>
<p>I’m thinking it’s both D and E as well… It changes concavity because f’‘(x) goes from positive y-values to negative y-values over the interval (0,2). At the point (1,0), f’'(x) = 0, therefore creating an inflection point.</p>
<p>If this were the AP and you had to guess, I would go with E.</p>
![http://i114.■■■■■■■■■■■■■■■/albums/n251/SignificaFire/CCI02212010_00000.jpg[/url]](http://i114.photobucket.com/albums/n251/SignificaFire/CCI02212010_00000.jpg%5B/url%5D){kind=link}
